Exercise 1: Radius first, then each endpoint with its own test

A power series centred at $a$ is a series $\sum_{n=0}^\infty c_n(x - a)^n$ : for each fixed $x$ it is an ordinary numerical series, which converges or not. The set of $x$ where it converges is always an interval centred at $a$ : there is a number $R$ , the radius of convergence, such that the series converges absolutely for $|x - a| < R$ and diverges for $|x - a| > R$ . At the two points $|x - a| = R$ , anything can happen.

The method has two stages. First the ratio test (or the root test) applied to $|a_n|$ with $x$ fixed, which gives $R$ . Then each endpoint, substituted into the series, which becomes a numerical series settled by the tests of the previous chapters.

a) Find the radius and the interval of convergence of $\sum_{n=1}^\infty \frac{x^n}{n\,3^n}$ .
b) Same question for $\sum_{n=1}^\infty \frac{n\,x^n}{4^n}$ .
c) Same question for $\sum_{n=0}^\infty \frac{(-1)^n x^n}{\sqrt{n + 1}}$ .
d) Same question for $\sum_{n=1}^\infty \frac{x^n}{n^2\,5^n}$ .
e) Explain why the ratio test can never decide an endpoint, and name the test that decided each endpoint in a) to d).

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Answers

a) $R = 3$ , interval $[-3, 3)$
b) $R = 4$ , interval $(-4, 4)$
c) $R = 1$ , interval $(-1, 1]$
d) $R = 5$ , interval $[-5, 5]$
e) At $|x - a| = R$ the ratio limit is $1$ by construction; endpoints: harmonic and AST (a), divergence test (b), AST and $p$ -series (c), absolute convergence with $p = 2$ (d).

a) Write $a_n = \frac{x^n}{n\,3^n}$ , with $x$ fixed and $x \ne 0$ . Ratio test on the absolute values: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|^{n+1}}{(n + 1)3^{n+1}} \cdot \frac{n\,3^n}{|x|^n} = \frac{|x|}{3} \cdot \frac{n}{n + 1} \to \frac{|x|}{3}$ . The series converges absolutely when $\frac{|x|}{3} < 1$ , that is $|x| < 3$ , and diverges when $|x| > 3$ : $R = 3$ . At $|x| = 3$ the limit is exactly $1$ , the case where the ratio test is silent, so each endpoint is a new series. At $x = 3$ : $\sum \frac{3^n}{n\,3^n} = \sum \frac{1}{n}$ , the harmonic series, which DIVERGES ( $p$ -series with $p = 1$ ). At $x = -3$ : $\sum \frac{(-3)^n}{n\,3^n} = \sum \frac{(-1)^n}{n}$ ; alternating series test: $b_n = \frac{1}{n}$ is decreasing and tends to $0$ , so it CONVERGES, conditionally since $\sum \frac{1}{n}$ diverges. Interval: $[-3, 3)$ .

b) $a_n = \frac{n\,x^n}{4^n}$ : $\left|\frac{a_{n+1}}{a_n}\right| = \frac{n + 1}{n} \cdot \frac{|x|}{4} \to \frac{|x|}{4}$ , so $R = 4$ . At $x = 4$ the term is $\frac{n\,4^n}{4^n} = n$ ; at $x = -4$ it is $(-1)^n n$ . In both cases $|a_n| = n \to \infty$ : the terms do not tend to $0$ and the series DIVERGES by the divergence test. Interval: $(-4, 4)$ . The divergence test is the first thing to try at an endpoint: when it applies, one line settles the question.

c) $a_n = \frac{(-1)^n x^n}{\sqrt{n + 1}}$ , $n \ge 0$ : $\left|\frac{a_{n+1}}{a_n}\right| = |x|\sqrt{\frac{n + 1}{n + 2}} \to |x|$ , so $R = 1$ . The sign $(-1)^n$ vanishes in the absolute value and plays no role in the radius, but it decides the endpoints. At $x = 1$ : $\sum \frac{(-1)^n}{\sqrt{n + 1}}$ is alternating with $b_n = \frac{1}{\sqrt{n + 1}}$ decreasing to $0$ : CONVERGES. At $x = -1$ : $(-1)^n(-1)^n = 1$ and the series is $\sum_{n\ge0} \frac{1}{\sqrt{n + 1}} = \sum_{k\ge1} \frac{1}{\sqrt{k}}$ , a $p$ -series with $p = \frac{1}{2} \le 1$ : DIVERGES. Interval: $(-1, 1]$ , the mirror image of a). The closed end is always on the side where the terms end up alternating.

d) $a_n = \frac{x^n}{n^2\,5^n}$ : $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{5} \cdot \frac{n^2}{(n + 1)^2} \to \frac{|x|}{5}$ , $R = 5$ . At $x = \pm 5$ : $|a_n| = \frac{5^n}{n^2\,5^n} = \frac{1}{n^2}$ , and $\sum \frac{1}{n^2}$ converges ( $p = 2 > 1$ ). So the series converges ABSOLUTELY at both endpoints, in a single argument that covers both signs. Interval: $[-5, 5]$ .

e) The radius is DEFINED by the ratio limit $\frac{|x|}{R}$ (or $\frac{|x - a|}{R}$ ) being less than $1$ ; at an endpoint that limit equals $1$ , and $L = 1$ is exactly the case where the ratio test says nothing. The same holds for the root test. So the endpoints are settled with the tests of the previous chapters: in a), the $p$ -series test ( $x = 3$ ) and the alternating series test ( $x = -3$ ); in b), the divergence test at both ends; in c), the alternating series test ( $x = 1$ ) and the $p$ -series test ( $x = -1$ ); in d), absolute convergence by the $p$ -series test at both ends. The four answers are the four possible shapes of an interval of convergence, drawn in the solution figure. A copy that writes $(-3, 3)$ for a) has the radius and loses the endpoint marks, usually two points of the question on a final.

Exercise 2: Centre, gaps and extreme radii: read the series before testing it

The centre is the number $a$ in the powers $(x - a)^n$ , and it must be read after the coefficient of $x$ has been factored out of the power. The radius is always obtained by applying the ratio or the root test to the TERMS $a_n$ , never by a formula on the coefficients when some powers are missing. Two extreme cases complete the picture: $R = 0$ (the series converges at its centre only) and $R = \infty$ (it converges for every real $x$ ).

a) Find the centre, the radius and the interval of convergence of $\sum_{n=1}^\infty \frac{(2x - 3)^n}{n}$ .
b) Same question for $\sum_{n=0}^\infty \frac{(x - 4)^{2n}}{9^n}$ . Why does the formula $R = \lim\left|\frac{c_n}{c_{n+1}}\right|$ with $c_n = \frac{1}{9^n}$ give a wrong radius here?
c) Where does $\sum_{n=0}^\infty n!\,(x - 1)^n$ converge?
d) Where does $\sum_{n=0}^\infty \frac{(x + 2)^n}{n!}$ converge?
e) Find the interval of convergence of $\sum_{n=1}^\infty \left(\frac{2n + 1}{n + 3}\right)^n (x + 1)^n$ .

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a) Centre $\frac{3}{2}$ , $R = \frac{1}{2}$ , interval $[1, 2)$
b) Centre $4$ , $R = 3$ , interval $(1, 7)$ ; only even powers appear, so the coefficient formula (which gives $9$ ) does not apply.
c) $R = 0$ : only at $x = 1$
d) $R = \infty$ : for every real $x$
e) Centre $-1$ , $R = \frac{1}{2}$ , interval $\left(-\frac{3}{2}, -\frac{1}{2}\right)$

a) Factor the coefficient of $x$ out of the power FIRST: $(2x - 3)^n = 2^n\left(x - \frac{3}{2}\right)^n$ , so the series is $\sum \frac{2^n}{n}\left(x - \frac{3}{2}\right)^n$ , centred at $a = \frac{3}{2}$ . Ratio test: $\left|\frac{a_{n+1}}{a_n}\right| = |2x - 3| \cdot \frac{n}{n + 1} \to |2x - 3|$ , and $|2x - 3| < 1 \iff \left|x - \frac{3}{2}\right| < \frac{1}{2}$ . Centre $\frac{3}{2}$ , radius $R = \frac{1}{2}$ , open interval $(1, 2)$ . At $x = 2$ : $2x - 3 = 1$ and the series is $\sum \frac{1}{n}$ , which DIVERGES. At $x = 1$ : $2x - 3 = -1$ and the series is $\sum \frac{(-1)^n}{n}$ , which CONVERGES by the alternating series test. Interval: $[1, 2)$ . The two classic misreadings are the centre $3$ (read off $-3$ without factoring) and the radius $1$ (read off $|2x - 3| < 1$ without dividing by $2$ ). The check: the midpoint of the interval found must be the centre, and its half-length the radius.

b) Only EVEN powers of $(x - 4)$ appear: the coefficients of the odd powers are $0$ , so there is no ratio $\frac{c_n}{c_{n+1}}$ between consecutive powers to take, and the shortcut built on $c_n = \frac{1}{9^n}$ would give $R = 9$ , which is false. Apply the ratio test to the terms: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x - 4|^{2n+2}}{9^{n+1}} \cdot \frac{9^n}{|x - 4|^{2n}} = \frac{(x - 4)^2}{9}$ , which is $< 1$ iff $|x - 4| < 3$ . Centre $4$ , $R = 3$ . At $x = 1$ and $x = 7$ : $(x - 4)^{2n} = 9^n$ , every term equals $1$ , the series DIVERGES. Interval: $(1, 7)$ . In fact this is a geometric series with ratio $r = \frac{(x - 4)^2}{9}$ , whose sum on $(1, 7)$ is $\frac{1}{1 - r} = \frac{9}{9 - (x - 4)^2}$ ; a geometric series never converges at its endpoints, where $|r| = 1$ .

c) $\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n + 1)!\,|x - 1|^{n+1}}{n!\,|x - 1|^n} = (n + 1)|x - 1|$ . For $x \ne 1$ this tends to $\infty$ , so the series diverges. At $x = 1$ every term with $n \ge 1$ is $0$ , and the series converges to its first term $0! = 1$ . Radius $R = 0$ : the set of convergence is the single point $\{1\}$ . A power series always converges at its centre; $R = 0$ means that is all it does.

d) $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x + 2|^{n+1}}{(n + 1)!} \cdot \frac{n!}{|x + 2|^n} = \frac{|x + 2|}{n + 1} \to 0 < 1$ for EVERY real $x$ . The series converges absolutely everywhere: $R = \infty$ , interval $(-\infty, \infty)$ , and there are no endpoints to test. Which function this series represents is the subject of the next chapter; here only its domain is asked. Compare with c): a factorial in the numerator kills the radius, one in the denominator makes it infinite.

e) Every factor is raised to the power $n$ , so the root test is the natural tool: $\sqrt[n]{|a_n|} = \frac{2n + 1}{n + 3}|x + 1| \to 2|x + 1|$ . Convergence iff $2|x + 1| < 1$ : centre $-1$ , $R = \frac{1}{2}$ , open interval $\left(-\frac{3}{2}, -\frac{1}{2}\right)$ . At the endpoints the root test gives exactly $1$ and is silent, like the ratio test in Exercise 1. At $x = -\frac{1}{2}$ , $x + 1 = \frac{1}{2}$ and $a_n = \left(\frac{2n + 1}{2(n + 3)}\right)^n = \left(1 - \frac{5}{2n + 6}\right)^n$ . The base tends to $1$ and the exponent to $\infty$ : the form $1^\infty$ is indeterminate. Take the logarithm, $n\ln\left(1 - \frac{5}{2n + 6}\right)$ , and treat it as in the chapter on sequences (L'Hôpital's rule on the quotient $\frac{\ln(1 - 5/(2x + 6))}{1/x}$ ): it tends to $-\frac{5}{2}$ . So $a_n \to e^{-5/2} \ne 0$ and the series DIVERGES by the divergence test. At $x = -\frac{3}{2}$ : $a_n = (-1)^n\left(1 - \frac{5}{2n + 6}\right)^n$ , $|a_n| \to e^{-5/2} \ne 0$ : DIVERGES. Interval: $\left(-\frac{3}{2}, -\frac{1}{2}\right)$ . The tempting shortcut 'the base tends to $1$ , so $a_n \to 1$ ' reaches the right verdict for a false reason, and loses the mark.

Exercise 3: Building series from 1/(1 - u): substitute, factor, recentre

The whole chapter rests on one formula, $\frac{1}{1 - u} = \sum_{n=0}^\infty u^n$ for $|u| < 1$ , and on the fact that any expression can be put in the place of $u$ . Put the function in the form $\frac{\text{constant}}{1 - u}$ , read off $u$ , and the condition $|u| < 1$ gives the interval directly, with no ratio test needed.

The figure shows the graph of $f(x) = \frac{3}{x^2 + x - 2}$ with its vertical asymptotes; the dot marks the centre $a = 0$ .

a) Find a power series centred at $0$ for $\frac{1}{1 + 3x}$ , and its interval of convergence.
b) Same question for $\frac{x^2}{4 - x}$ .
c) Find a power series centred at $a = 2$ for $\frac{1}{x + 5}$ , and its interval of convergence.
d) Using partial fractions, find a power series centred at $0$ for $f(x) = \frac{3}{x^2 + x - 2}$ . Give its first three coefficients and its radius.
e) Read the radius of d) on the figure. Predict the radius of the series of $f$ centred at $a = 3$ , then at $a = -3$ , and check the first prediction.

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a) $\sum_{n=0}^\infty (-3)^n x^n$ , interval $\left(-\frac{1}{3}, \frac{1}{3}\right)$
b) $\sum_{n=0}^\infty \frac{x^{n+2}}{4^{n+1}}$ , interval $(-4, 4)$
c) $\sum_{n=0}^\infty \frac{(-1)^n (x - 2)^n}{7^{n+1}}$ , interval $(-5, 9)$
d) $\sum_{n=0}^\infty \left(-1 - \frac{(-1)^n}{2^{n+1}}\right)x^n$ ; $c_0 = -\frac{3}{2}$ , $c_1 = -\frac{3}{4}$ , $c_2 = -\frac{9}{8}$ ; $R = 1$
e) $R$ is the distance from the centre to the nearest asymptote: $1$ at $a = 0$ , $2$ at $a = 3$ , $1$ at $a = -3$ .

a) $\frac{1}{1 + 3x} = \frac{1}{1 - (-3x)}$ , so $u = -3x$ and $\frac{1}{1 + 3x} = \sum_{n=0}^\infty (-3x)^n = \sum_{n=0}^\infty (-3)^n x^n$ , valid for $|-3x| < 1$ , that is $|x| < \frac{1}{3}$ . At $x = \pm\frac{1}{3}$ , $u = \mp 1$ : the terms $u^n$ have absolute value $1$ and do not tend to $0$ , so the series diverges; a geometric series never converges where $|u| = 1$ . Interval: $\left(-\frac{1}{3}, \frac{1}{3}\right)$ . Check at $x = 0$ : the series gives its first term $1$ , and $\frac{1}{1 + 0} = 1$ .

b) Force the $1$ in the denominator: $\frac{x^2}{4 - x} = \frac{x^2}{4} \cdot \frac{1}{1 - \frac{x}{4}} = \frac{x^2}{4}\sum_{n=0}^\infty \frac{x^n}{4^n} = \sum_{n=0}^\infty \frac{x^{n+2}}{4^{n+1}}$ , valid for $\left|\frac{x}{4}\right| < 1$ . At $x = 4$ every term equals $\frac{4^{n+2}}{4^{n+1}} = 4$ , and at $x = -4$ every term is $\pm 4$ : divergence at both. Interval $(-4, 4)$ . Multiplying by $x^2$ shifts every power by two and changes nothing else. With $k = n + 2$ the series is also $\sum_{k=2}^\infty \frac{x^k}{4^{k-1}}$ ; both forms are right as long as the first term is $\frac{x^2}{4}$ . The frequent error is $\frac{1}{4 - x} = 4 \cdot \frac{1}{1 - x/4}$ : the $4$ comes out as $\frac{1}{4}$ , and the slip multiplies everything by $16$ .

c) Centred at $2$ means powers of $(x - 2)$ , so the denominator is rewritten around $2$ : $x + 5 = 7 + (x - 2) = 7\left(1 + \frac{x - 2}{7}\right)$ . Then $\frac{1}{x + 5} = \frac{1}{7} \cdot \frac{1}{1 - \left(-\frac{x - 2}{7}\right)} = \sum_{n=0}^\infty \frac{(-1)^n (x - 2)^n}{7^{n+1}}$ , valid for $\left|\frac{x - 2}{7}\right| < 1$ , that is $|x - 2| < 7$ : interval $(-5, 9)$ , endpoints excluded as for every geometric series. The radius $7$ is the distance from the centre $2$ to $-5$ , where $\frac{1}{x + 5}$ blows up. Check at $x = 2$ : the series gives $\frac{1}{7} = \frac{1}{2 + 5}$ .

d) Partial fractions first: $x^2 + x - 2 = (x - 1)(x + 2)$ and $\frac{3}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}$ with $3 = A(x + 2) + B(x - 1)$ ; $x = 1$ gives $A = 1$ and $x = -2$ gives $B = -1$ . Each piece is geometric: $\frac{1}{x - 1} = -\frac{1}{1 - x} = -\sum x^n$ for $|x| < 1$ , and $\frac{1}{x + 2} = \frac{1}{2} \cdot \frac{1}{1 + x/2} = \sum \frac{(-1)^n x^n}{2^{n+1}}$ for $|x| < 2$ . Subtracting, $f(x) = \sum_{n=0}^\infty \left(-1 - \frac{(-1)^n}{2^{n+1}}\right)x^n$ , valid where BOTH series converge: $|x| < 1$ . First coefficients: $c_0 = -1 - \frac{1}{2} = -\frac{3}{2}$ , $c_1 = -1 + \frac{1}{4} = -\frac{3}{4}$ , $c_2 = -1 - \frac{1}{8} = -\frac{9}{8}$ . Check: $f(0) = \frac{3}{-2} = c_0$ . The radius is exactly $1$ , not more: $c_n \to -1$ , so at $x = \pm 1$ the terms do not tend to $0$ and the ratio of consecutive coefficients tends to $1$ . $R = 1$ , interval $(-1, 1)$ .

e) On the figure the asymptotes are $x = -2$ and $x = 1$ , and the centre $0$ is at distance $1$ from the nearer one: $R = 1$ is exactly that distance. It cannot be larger: a power series is continuous on its open interval, so it cannot equal $f$ on an interval that contains $x = 1$ , where $f$ is unbounded. About $a = 3$ the nearer asymptote is $x = 1$ , at distance $2$ : predicted $R = 2$ . About $a = -3$ the nearer is $x = -2$ , at distance $1$ : predicted $R = 1$ . Check of the first, as in d): $\frac{1}{x - 1} = \frac{1}{2 + (x - 3)}$ has radius $2$ and $\frac{1}{x + 2} = \frac{1}{5 + (x - 3)}$ has radius $5$ ; the difference converges where both do, so $R = 2$ . Keep this reading for rational functions and as a check only: Exercise 4 shows a function with no vertical asymptote at all whose series still has radius $1$ .

Exercise 4: Partial sums against the function: inside, at and beyond the radius

Replacing $u$ by $-x^2$ in the geometric series gives a series for $f(x) = \frac{1}{1 + x^2}$ , a function that is defined, continuous and bounded on the whole real line. Write $S_N(x)$ for the partial sum of the series up to the term in $x^{2N}$ .

The figure shows $f$ with $S_2(x) = 1 - x^2 + x^4$ and $S_3(x) = 1 - x^2 + x^4 - x^6$ on $[-1.6, 1.6]$ ; the dashed lines are $x = \pm 1$ .

a) Find the power series of $f$ centred at $0$ and its interval of convergence.
b) Show that $f(x) - S_N(x) = \frac{(-x^2)^{N+1}}{1 + x^2}$ for every real $x$ . Compute $S_3\left(\frac{1}{2}\right)$ and $f\left(\frac{1}{2}\right) - S_3\left(\frac{1}{2}\right)$ exactly.
c) Compute $S_0(2)$ , $S_1(2)$ , $S_2(2)$ and $S_3(2)$ . Does the series converge at $x = 2$ , where $f(2) = \frac{1}{5}$ ? Describe what the partial sums do on the figure for $|x| > 1$ .
d) At $x = 1$ , a student notices that the partial sums average $\frac{1}{2} = f(1)$ and concludes that the series equals $f$ there. Answer him, then explain why the radius is $1$ although $f$ has no vertical asymptote.
e) Using b), find the smallest $N$ for which $S_N\left(\frac{1}{2}\right)$ approximates $f\left(\frac{1}{2}\right)$ within $\frac{1}{1000}$ .

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a) $\sum_{n=0}^\infty (-1)^n x^{2n}$ , interval $(-1, 1)$
b) $S_3\left(\frac{1}{2}\right) = \frac{51}{64}$ , error $\frac{1}{320}$
c) $1, -3, 13, -51$ : diverges, although $f(2) = \frac{1}{5}$
d) The partial sums $1, 0, 1, 0, \dots$ have no limit: the series diverges at $1$ . The radius comes from $|x^2| < 1$ , not from the graph.
e) $N = 4$ (error $\frac{1}{1280}$ )

a) $\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots$ , valid for $|-x^2| < 1$ , that is $|x| < 1$ . At $x = \pm 1$ the terms are $(-1)^n$ , which do not tend to $0$ : diverges. Interval: $(-1, 1)$ , although $f$ itself is defined everywhere.

b) $S_N(x)$ is a FINITE geometric sum with ratio $u = -x^2$ , so $S_N = \frac{1 - u^{N+1}}{1 - u} = \frac{1 - (-x^2)^{N+1}}{1 + x^2}$ for every real $x$ : a finite sum has no convergence condition. Hence $f(x) - S_N(x) = \frac{1}{1 + x^2} - \frac{1 - (-x^2)^{N+1}}{1 + x^2} = \frac{(-x^2)^{N+1}}{1 + x^2}$ . At $x = \frac{1}{2}$ : $S_3 = 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} = \frac{64 - 16 + 4 - 1}{64} = \frac{51}{64}$ , $f\left(\frac{1}{2}\right) = \frac{1}{5/4} = \frac{4}{5}$ , and $f - S_3 = \frac{(1/4)^4}{5/4} = \frac{1}{256} \cdot \frac{4}{5} = \frac{1}{320}$ ; directly, $\frac{4}{5} - \frac{51}{64} = \frac{256 - 255}{320} = \frac{1}{320}$ . The formula gives the error EXACTLY, a luxury that only geometric series offer.

c) By b), $S_N(2) = \frac{1 - (-4)^{N+1}}{5}$ : $S_0 = 1$ , $S_1 = 1 - 4 = -3$ , $S_2 = -3 + 16 = 13$ , $S_3 = 13 - 64 = -51$ . The partial sums swing with growing amplitude; the terms $(-4)^n$ do not tend to $0$ , so the series DIVERGES at $x = 2$ , even though $f(2) = \frac{1}{5}$ is a perfectly good number. On the figure, as soon as $|x| > 1$ , $S_2$ shoots up and $S_3$ plunges, on both sides: the partial sums alternate above and below $f$ with a gap $\frac{x^{2N+2}}{1 + x^2}$ that GROWS with $N$ . Inside $(-1, 1)$ the same gap shrinks to $0$ , which is why all the curves hug $f$ near $0$ .

d) $S_N(1) = \frac{1 - (-1)^{N+1}}{2}$ gives $1, 0, 1, 0, \dots$ : the sequence of partial sums has no limit, so the series DIVERGES at $x = 1$ and represents nothing there. The average of $1$ and $0$ is indeed $\frac{1}{2} = f(1)$ , but the sum of a series is the LIMIT of its partial sums, not their average; the student's argument would also make $1 - 1 + 1 - \cdots$ equal to $\frac{1}{2}$ , which the divergence test rules out. As for the radius, nothing on the real graph explains it, since $f$ is smooth everywhere: the obstruction is the condition $|x^2| < 1$ under which the series was built, and that condition is the whole story. (The deeper reason is that $1 + x^2$ vanishes at the complex numbers $\pm i$ , at distance $1$ from $0$ ; this lies outside MATH 141.) The lesson for the exam: a radius is COMPUTED, never read off a graph.

e) By b), $\left|f\left(\frac{1}{2}\right) - S_N\left(\frac{1}{2}\right)\right| = \frac{(1/4)^{N+1}}{5/4} = \frac{4}{5} \cdot \frac{1}{4^{N+1}}$ . We need $\frac{4}{5 \cdot 4^{N+1}} < \frac{1}{1000}$ , that is $4^{N+1} > 800$ . Since $4^4 = 256 < 800 < 1024 = 4^5$ , the smallest choice is $N + 1 = 5$ , so $N = 4$ : the five terms up to $x^8$ , with error $\frac{4}{5 \cdot 1024} = \frac{1}{1280}$ . For $N = 3$ the error is $\frac{1}{320}$ , too large. No calculator is needed because the error formula is exact; error bounds for series that are not geometric come later in the course.

Exercise 5: Differentiate term by term, then evaluate: 1/(1 - x) squared and its cousins

Theorem (term-by-term calculus). If $\sum c_n(x - a)^n$ has radius $R > 0$ , its sum $f$ is differentiable on $(a - R, a + R)$ , with $f'(x) = \sum_{n=1}^\infty n c_n (x - a)^{n-1}$ and $\int f(x)\,dx = C + \sum_{n=0}^\infty c_n\frac{(x - a)^{n+1}}{n + 1}$ , and both series have the SAME radius $R$ . The theorem says nothing about the endpoints.

Differentiating a known series is the standard way to sum numerical series whose terms carry a factor $n$ or $n^2$ .

a) Differentiate the geometric series to obtain a power series for $\frac{1}{(1 - x)^2}$ written with $x^n$ , and give its interval of convergence.
b) Deduce a series for $\frac{x}{(1 - x)^2}$ , then the sums $\sum_{n=1}^\infty \frac{n}{2^n}$ and $\sum_{n=1}^\infty \frac{n}{3^n}$ .
c) Differentiate once more to find $\sum_{n=1}^\infty n^2 x^n$ in closed form on $(-1, 1)$ , then $\sum_{n=1}^\infty \frac{n^2}{2^n}$ .
d) Evaluate $\sum_{n=0}^\infty \frac{(-1)^n (n + 1)}{4^n}$ .
e) A student writes: $\sum_{n=0}^\infty \left(\frac{1}{2}\right)^n = 2$ , and differentiating both sides gives $\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n-1} = 0$ . Find the error and the correct value.

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a) $\frac{1}{(1 - x)^2} = \sum_{n=0}^\infty (n + 1)x^n$ on $(-1, 1)$
b) $\frac{x}{(1 - x)^2} = \sum_{n=1}^\infty n x^n$ ; the sums are $2$ and $\frac{3}{4}$ .
c) $\sum_{n=1}^\infty n^2 x^n = \frac{x(1 + x)}{(1 - x)^3}$ ; $\sum \frac{n^2}{2^n} = 6$
d) $\frac{16}{25}$
e) A number was differentiated instead of a function; correct value $4$ .

a) On $(-1, 1)$ , $\frac{1}{1 - x} = \sum_{n=0}^\infty x^n$ . Differentiate both sides. On the left, the derivative of $(1 - x)^{-1}$ is $(1 - x)^{-2}$ by the chain rule (two minus signs cancel). On the right, $\sum_{n=1}^\infty n x^{n-1}$ : the $n = 0$ term is the constant $1$ , whose derivative is $0$ , hence the start at $n = 1$ . Shift the index with $k = n - 1$ : $\frac{1}{(1 - x)^2} = \sum_{k=0}^\infty (k + 1)x^k = 1 + 2x + 3x^2 + \cdots$ . The theorem gives the radius $1$ ; the endpoints are checked by hand: at $x = \pm 1$ , $|(k + 1)x^k| = k + 1 \to \infty$ , diverges. Interval $(-1, 1)$ . Check at $x = 0$ : $1 = \frac{1}{1^2}$ .

b) Multiply a) by $x$ (in its form $\sum_{n\ge1} n x^{n-1}$ ): $\frac{x}{(1 - x)^2} = \sum_{n=1}^\infty n x^n$ on $(-1, 1)$ . At $x = \frac{1}{2}$ , inside the interval: $\sum \frac{n}{2^n} = \frac{1/2}{(1/2)^2} = 2$ . At $x = \frac{1}{3}$ : $\sum \frac{n}{3^n} = \frac{1/3}{(2/3)^2} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$ . The solution figure shows the partial sums of the first series climbing to $2$ : $\frac{1}{2}$ , $1$ , $\frac{11}{8}$ , $\frac{13}{8}$ , and so on.

c) Differentiate $\sum_{n\ge1} n x^n = \frac{x}{(1 - x)^2}$ by the quotient rule: $\sum_{n\ge1} n^2 x^{n-1} = \frac{(1 - x)^2 + 2x(1 - x)}{(1 - x)^4} = \frac{(1 - x) + 2x}{(1 - x)^3} = \frac{1 + x}{(1 - x)^3}$ , after cancelling one factor $1 - x$ . Multiply by $x$ : $\sum_{n\ge1} n^2 x^n = \frac{x(1 + x)}{(1 - x)^3}$ for $|x| < 1$ . At $x = \frac{1}{2}$ : $\frac{\frac{1}{2} \cdot \frac{3}{2}}{\frac{1}{8}} = 6$ . The pattern 'differentiate, then multiply by $x$ ' restores the power $x^n$ and brings down one more factor $n$ each time; it sums $\sum n^k x^n$ for any $k$ .

d) It is a) at $x = -\frac{1}{4}$ , which lies in $(-1, 1)$ : the general term $\frac{(-1)^n(n + 1)}{4^n}$ is $(n + 1)x^n$ with $x = -\frac{1}{4}$ , sign included. So the sum is $\frac{1}{(1 + 1/4)^2} = \frac{1}{25/16} = \frac{16}{25}$ . Recognising a series means matching its general term to a known one AND naming the value of $x$ , then checking that it lies in the open interval.

e) The sum $2$ is a NUMBER, the value of $\frac{1}{1 - x}$ at one point; the derivative of a constant is $0$ and carries no information about the function. Differentiation acts on functions of $x$ : differentiate $\frac{1}{1 - x} = \sum x^n$ on $(-1, 1)$ first, then evaluate: $\sum_{n\ge1} n\left(\frac{1}{2}\right)^{n-1} = \frac{1}{(1 - 1/2)^2} = 4$ . Consistent with b): $\sum \frac{n}{2^n} = \frac{1}{2}\sum n\left(\frac{1}{2}\right)^{n-1} = \frac{1}{2} \cdot 4 = 2$ . The rule: keep $x$ a variable until the last line, and substitute the number only once the identity between functions is established.

Exercise 6: Integrate term by term: ln(1 + x), arctan x and the constant

Integrating a geometric series term by term produces the two series of the chapter that are not rational functions. The theorem of Exercise 5 keeps the radius; the constant of integration and the endpoints are your job. Admit also Abel's theorem: if a power series converges at an endpoint of its interval and the function it represents is continuous there, the equality between the function and the series holds at that endpoint too.

The figure shows $y = \ln(1 + x)$ with the partial sums $S_5$ and $S_6$ of its power series, the sums of its terms up to $x^5$ and $x^6$ .

a) Integrate the series of $\frac{1}{1 + t}$ from $0$ to $x$ to obtain the power series of $\ln(1 + x)$ , justifying the constant.
b) Find the interval of convergence of this series, testing each endpoint, and compare it with the interval of the series of $\frac{1}{1 + x}$ . Deduce the sum of $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ , and say what the figure shows beyond $x = 1$ .
c) Same work for $\arctan x$ , starting from $\frac{1}{1 + t^2}$ . Deduce the sum of $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$ .
d) Find the power series of $\ln(5 - x)$ centred at $0$ , and its interval of convergence.
e) Write $\int_0^{1/2} \frac{dx}{1 + x^4}$ as the sum of a numerical series. Show that its first three terms give the integral within $10^{-5}$ , and give that approximation as an exact fraction.

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Answers

a) $\ln(1 + x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$ , constant $\ln 1 = 0$
b) Interval $(-1, 1]$ : the endpoint $1$ is gained; $1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2$ .
c) $\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n + 1}$ on $[-1, 1]$ ; $1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}$
d) $\ln(5 - x) = \ln 5 - \sum_{n=1}^\infty \frac{x^n}{n\,5^n}$ , interval $[-5, 5)$
e) $\sum_{n=0}^\infty \frac{(-1)^n}{(4n + 1)2^{4n+1}}$ ; $\frac{1}{2} - \frac{1}{160} + \frac{1}{4608} = \frac{11381}{23040}$ , error below $\frac{1}{106496}$

a) On $(-1, 1)$ , $\frac{1}{1 + t} = \frac{1}{1 - (-t)} = \sum_{n=0}^\infty (-1)^n t^n$ . For $|x| < 1$ the interval $[0, x]$ lies inside the interval of convergence, so the series can be integrated term by term on it: $\ln(1 + x) = \int_0^x \frac{dt}{1 + t} = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n + 1} = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ . The definite integral from $0$ fixes the constant by itself. With an indefinite integral one writes $\ln(1 + x) = C + \sum(\cdots)$ and finds $C$ at the centre: $\ln 1 = C + 0$ , so $C = 0$ . The constant is $0$ here only because $\ln(1 + 0) = 0$ ; d) shows a case where it is not.

b) Radius $1$ by the theorem (or by the ratio test: $\frac{n}{n + 1}|x| \to |x|$ ). At $x = 1$ : $\sum \frac{(-1)^{n+1}}{n}$ , the alternating harmonic series, converges by the alternating series test, since $\frac{1}{n}$ decreases to $0$ . At $x = -1$ : $(-1)^{n+1}(-1)^n = -1$ , so the series is $-\sum \frac{1}{n}$ , which diverges. Interval: $(-1, 1]$ . The series of $\frac{1}{1 + x}$ had interval $(-1, 1)$ : integrating kept $R$ and GAINED the endpoint $x = 1$ . Since $\ln(1 + x)$ is continuous at $1$ and the series converges there, Abel's theorem gives $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2$ . Beyond $x = 1$ , the figure shows $S_5$ and $S_6$ leaving $\ln(1 + x)$ in opposite directions: for $x > 1$ the terms $\frac{x^n}{n}$ grow without bound, and the series represents nothing there, although $\ln(1 + x)$ exists. Near $x = -1$ , $\ln(1 + x) \to -\infty$ while every partial sum stays finite: the divergence at $-1$ is visible as well.

c) $\frac{1}{1 + t^2} = \sum_{n=0}^\infty (-1)^n t^{2n}$ for $|t| < 1$ (Exercise 4). Integrate from $0$ to $x$ : $\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n + 1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ , with constant $\arctan 0 = 0$ . At $x = 1$ : $\sum \frac{(-1)^n}{2n + 1}$ , alternating with $\frac{1}{2n + 1}$ decreasing to $0$ , converges. At $x = -1$ : $(-1)^{2n+1} = -1$ , so the series is $-\sum \frac{(-1)^n}{2n + 1}$ , which converges as well. Interval $[-1, 1]$ : BOTH endpoints are gained, where $\frac{1}{1 + x^2}$ had $(-1, 1)$ . Both converge only conditionally, since $\sum \frac{1}{2n + 1}$ diverges by limit comparison with $\sum \frac{1}{n}$ . Abel's theorem at $x = 1$ : $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \arctan 1 = \frac{\pi}{4}$ .

d) Differentiate first: $\frac{d}{dx}\ln(5 - x) = -\frac{1}{5 - x} = -\frac{1}{5} \cdot \frac{1}{1 - \frac{x}{5}} = -\sum_{n=0}^\infty \frac{x^n}{5^{n+1}}$ for $|x| < 5$ . Integrate term by term: $\ln(5 - x) = C - \sum_{n=0}^\infty \frac{x^{n+1}}{(n + 1)5^{n+1}} = C - \sum_{n=1}^\infty \frac{x^n}{n\,5^n}$ . At the centre $x = 0$ : $\ln 5 = C - 0$ , so $C = \ln 5$ and $\ln(5 - x) = \ln 5 - \sum_{n=1}^\infty \frac{x^n}{n\,5^n}$ . Forgetting $C$ gives a series worth $0$ at $x = 0$ where the function is worth $\ln 5$ : the check at the centre catches it in one line. Endpoints: at $x = 5$ , $\sum \frac{1}{n}$ diverges (and indeed $\ln(5 - x) \to -\infty$ as $x \to 5^-$ ); at $x = -5$ , $\sum \frac{(-1)^n}{n}$ converges by the alternating series test. Interval: $[-5, 5)$ .

e) For $|x| < 1$ , $\frac{1}{1 + x^4} = \sum_{n=0}^\infty (-1)^n x^{4n}$ ( $u = -x^4$ ). The interval of integration $[0, \frac{1}{2}]$ lies inside $(-1, 1)$ , so term-by-term integration is justified: $\int_0^{1/2} \frac{dx}{1 + x^4} = \sum_{n=0}^\infty (-1)^n\left[\frac{x^{4n+1}}{4n + 1}\right]_0^{1/2} = \sum_{n=0}^\infty \frac{(-1)^n}{(4n + 1)2^{4n+1}} = \frac{1}{2} - \frac{1}{160} + \frac{1}{4608} - \frac{1}{106496} + \cdots$ . This numerical series is alternating with $b_n = \frac{1}{(4n + 1)2^{4n+1}}$ decreasing to $0$ , so by the alternating series estimate the error after three terms is at most the fourth, $\frac{1}{13 \cdot 2^{13}} = \frac{1}{106496} < 10^{-5}$ . The approximation, over the common denominator $23040 = 2^9 \cdot 45$ : $\frac{1}{2} - \frac{1}{160} + \frac{1}{4608} = \frac{11520 - 144 + 5}{23040} = \frac{11381}{23040}$ , about $0.4940$ . The function has an elementary antiderivative, but it needs a four-factor partial fraction decomposition; the series gives the value to $10^{-5}$ in three lines.

Exercise 7: The same radius, not the same interval: endpoints lost and gained

The theorem of Exercise 5 promises that $f'$ and $\int f$ have the same RADIUS as $f$ . It promises nothing at the endpoints. Let $f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}$ and $F(x) = \int_0^x f(t)\,dt$ .

a) Find the interval of convergence of the series of $f$ .
b) Write $f'(x)$ as a power series and find its interval of convergence.
c) Same question for $f''(x)$ .
d) Same question for $F(x)$ . A student claims that differentiating or integrating a power series never changes its interval of convergence: correct him, and say what can happen at an endpoint under each operation.
e) Using the series of $\ln(1 - x)$ , show that $f'(x) = -\frac{\ln(1 - x)}{x}$ for $0 < |x| < 1$ , and check this formula against the value of the series of b) at $x = -1$ .

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Answers

a) $[-1, 1]$
b) $f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n}$ , interval $[-1, 1)$
c) $f''(x) = \sum_{n=2}^\infty \frac{(n - 1)x^{n-2}}{n}$ , interval $(-1, 1)$
d) $F(x) = \sum_{n=1}^\infty \frac{x^{n+1}}{n^2(n + 1)}$ , interval $[-1, 1]$ ; same $R$ , but differentiating can lose endpoints and integrating can gain them.
e) $-\frac{\ln(1 - x)}{x} = \sum \frac{x^{n-1}}{n}$ ; at $x = -1$ both give $\ln 2$ .

a) Ratio test: $\frac{n^2}{(n + 1)^2}|x| \to |x|$ , so $R = 1$ . At $x = \pm 1$ : $\left|\frac{(\pm 1)^n}{n^2}\right| = \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges ( $p = 2$ ): absolute convergence at both endpoints. Interval $[-1, 1]$ .

b) $f'(x) = \sum_{n=1}^\infty \frac{n x^{n-1}}{n^2} = \sum_{n=1}^\infty \frac{x^{n-1}}{n}$ , radius $1$ by the theorem. At $x = 1$ : $\sum \frac{1}{n}$ , which diverges. At $x = -1$ : $\sum \frac{(-1)^{n-1}}{n}$ , which converges by the alternating series test. Interval $[-1, 1)$ : the endpoint $x = 1$ is LOST.

c) The $n = 1$ term of $f'$ is the constant $1$ , so $f''(x) = \sum_{n=2}^\infty \frac{(n - 1)x^{n-2}}{n}$ . At $x = \pm 1$ : $\left|\frac{n - 1}{n}(\pm 1)^{n-2}\right| = \frac{n - 1}{n} \to 1 \ne 0$ , divergence at both ends by the divergence test. Interval $(-1, 1)$ : the endpoint $-1$ is lost in turn.

d) Integrating term by term from $0$ (constant $F(0) = 0$ ): $F(x) = \sum_{n=1}^\infty \frac{x^{n+1}}{n^2(n + 1)}$ , radius $1$ . At $x = \pm 1$ : $\left|\frac{(\pm 1)^{n+1}}{n^2(n + 1)}\right| \le \frac{1}{n^3}$ , absolute convergence by comparison with $p = 3$ . Interval $[-1, 1]$ . The student is right about the radius, which the theorem guarantees, and wrong about the interval: $[-1, 1]$ for $f$ , $[-1, 1)$ for $f'$ , $(-1, 1)$ for $f''$ , $[-1, 1]$ for $F$ , as drawn in the solution figure. Differentiating multiplies the $n$ th term by $n$ , which makes the terms at an endpoint LARGER: it can only lose an endpoint. Integrating divides by $n + 1$ : it can only gain one. (That these two directions hold for every power series is true, and admitted here.) On a copy, every derived or integrated series gets its own endpoint tests, even when its radius is taken from the theorem.

e) Exercise 6 with $-x$ in place of $x$ gives $\ln(1 - x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{(-x)^n}{n} = -\sum_{n=1}^\infty \frac{x^n}{n}$ for $-1 \le x < 1$ . For $0 < |x| < 1$ , divide by $-x$ : $-\frac{\ln(1 - x)}{x} = \sum_{n=1}^\infty \frac{x^{n-1}}{n} = f'(x)$ . At $x = 0$ the series gives $f'(0) = 1$ , which is the limit of $-\frac{\ln(1 - x)}{x}$ as $x \to 0$ by L'Hôpital's rule. At $x = -1$ the series of b) converges to $1 - \frac{1}{2} + \frac{1}{3} - \cdots = \ln 2$ (Exercise 6 b), and the formula gives $-\frac{\ln 2}{-1} = \ln 2$ : they agree, as Abel's theorem predicts. At $x = 1$ the formula involves $\ln 0$ , which does not exist, matching the lost endpoint. The function $f$ itself has no elementary closed form; its derivative does.

Exercise 8: Five statements to correct

Each statement below was written on a MATH 141 practice paper, and each one is false or incomplete. Say what is wrong, give the correct statement, and settle it with a short computation or a counterexample.

a) The series $\sum c_n x^n$ converges at $x = -4$ , so it also converges at $x = 4$ .
b) For $\sum \frac{x^n}{n^2\,3^n}$ , the ratio test gives the limit $\frac{|x|}{3} < 1$ , so the interval of convergence is $(-3, 3)$ .
c) $\frac{1}{2 + x} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{2^n}$ for $|x| < 1$ .
d) If $\sum c_n x^n$ has radius $2$ , then $\sum c_n x^{2n}$ has radius $4$ .
e) Since $\ln(1 + x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$ , putting $x = 2$ gives $\ln 3 = 2 - 2 + \frac{8}{3} - 4 + \cdots$ .

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Answers

a) False: $\sum \frac{x^n}{n\,4^n}$ converges at $-4$ , diverges at $4$ . Correct: convergence for every $|x| < 4$ .
b) Incomplete: both endpoints converge ( $p = 2$ ), the interval is $[-3, 3]$ .
c) False twice: $\frac{1}{2 + x} = \sum \frac{(-1)^n x^n}{2^{n+1}}$ for $|x| < 2$ .
d) False: the radius is $\sqrt{2}$ .
e) False: $x = 2$ is outside $(-1, 1]$ and the series diverges; correct: $\ln 3 = \sum \frac{(2/3)^n}{n}$ .

a) FALSE. Counterexample: $\sum \frac{x^n}{n\,4^n}$ , of radius $4$ . At $x = -4$ it is $\sum \frac{(-1)^n}{n}$ , convergent by the alternating series test; at $x = 4$ it is $\sum \frac{1}{n}$ , divergent. Correct statement: if $\sum c_n x^n$ converges at $x = -4$ , then $R \ge 4$ , so the series converges ABSOLUTELY for every $|x| < 4$ , for instance at $x = 3$ or at $x = -3.9$ ; at $x = 4$ nothing can be said. Symmetrically, divergence at $x = 6$ gives $R \le 6$ and divergence for every $|x| > 6$ .

b) INCOMPLETE, hence wrong as an answer. The ratio test proves convergence on $(-3, 3)$ and divergence for $|x| > 3$ ; at $x = \pm 3$ its limit is $1$ and it says nothing. Here $|a_n| = \frac{3^n}{n^2\,3^n} = \frac{1}{n^2}$ at both endpoints, and $\sum \frac{1}{n^2}$ converges ( $p = 2$ ): the series converges absolutely at $x = 3$ and at $x = -3$ . Correct statement: $R = 3$ and the interval of convergence is $[-3, 3]$ .

c) FALSE, twice. The centre test finds the first error: at $x = 0$ the series gives $1$ , the function $\frac{1}{2}$ . The $1$ of $\frac{1}{1 - u}$ must be created by factoring out the $2$ : $\frac{1}{2 + x} = \frac{1}{2} \cdot \frac{1}{1 + \frac{x}{2}} = \sum_{n=0}^\infty \frac{(-1)^n x^n}{2^{n+1}}$ , valid for $\left|\frac{x}{2}\right| < 1$ , that is $|x| < 2$ . The student's radius $1$ is the second error: the condition $|u| < 1$ applies to $u = -\frac{x}{2}$ , not to $x$ .

d) FALSE. The series $\sum c_n u^n$ converges for $|u| < 2$ and diverges for $|u| > 2$ . With $u = x^2$ , the series $\sum c_n x^{2n}$ converges for $x^2 < 2$ and diverges for $x^2 > 2$ : its radius is $\sqrt{2}$ , not $4$ . Example: $c_n = \frac{1}{2^n}$ gives $\sum \frac{x^{2n}}{2^n}$ , geometric with ratio $\frac{x^2}{2}$ , convergent iff $|x| < \sqrt{2}$ . Substituting $x^2$ takes the square root of the radius; substituting $3x$ divides it by $3$ .

e) FALSE. The identity holds only on $(-1, 1]$ , and $2$ is outside: at $x = 2$ the terms $\frac{2^n}{n}$ tend to $\infty$ , so the series diverges and equals nothing, let alone $\ln 3$ . Correct method: find a known series and a point INSIDE its interval that produce the value. With $\ln(1 - x) = -\sum \frac{x^n}{n}$ on $[-1, 1)$ and $x = \frac{2}{3}$ : $\ln 3 = -\ln\frac{1}{3} = -\ln\left(1 - \frac{2}{3}\right) = \sum_{n=1}^\infty \frac{(2/3)^n}{n} = \frac{2}{3} + \frac{2}{9} + \frac{8}{81} + \cdots$ . A formula proved on an interval says nothing outside it.

Exercise 9: Waiting for a six: expected values as power series

A board game asks a player to roll a fair die until the first six appears. Let $N$ be the number of rolls needed. The rolls are independent, so $N = n$ means $n - 1$ rolls without a six followed by a six: $P(N = n) = \frac{1}{6}\left(\frac{5}{6}\right)^{n-1}$ for $n = 1, 2, 3, \dots$ The expected value and the second moment are the numerical series $E(N) = \sum_{n\ge1} n\,P(N = n)$ and $E(N^2) = \sum_{n\ge1} n^2\,P(N = n)$ .

The figure shows the probabilities $P(N = n)$ for $n \le 14$ . No calculator: give exact answers.

a) Check that these probabilities add up to $1$ .
b) Compute $E(N)$ by recognising a power series of Exercise 5 evaluated at a point, and say why that point is allowed.
c) Compute $E(N^2)$ , then the variance $\text{Var}(N) = E(N^2) - E(N)^2$ .
d) Two players A and B roll in turn, A first, and the first to roll a six wins. Compute the probability that A wins.
e) For an attempt that succeeds with probability $p$ , $0 < p \le 1$ , show that $E(N) = \frac{1}{p}$ . What goes wrong when $p = 0$ , in terms of the interval of convergence? Apply the formula to waiting for a double six with two dice.

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Answers

a) Geometric, first term $\frac{1}{6}$ , ratio $\frac{5}{6}$ : sum $1$
b) $E(N) = \frac{1}{6} \cdot \frac{1}{(1 - 5/6)^2} = 6$
c) $E(N^2) = 66$ , $\text{Var}(N) = 30$
d) $P(\text{A wins}) = \frac{6}{11}$
e) $E(N) = p \cdot \frac{1}{p^2} = \frac{1}{p}$ ; $p = 0$ puts $x = 1$ at the excluded endpoint; double six: $36$ throws.

a) $\sum_{n=1}^\infty \frac{1}{6}\left(\frac{5}{6}\right)^{n-1}$ is geometric with first term $\frac{1}{6}$ and ratio $\frac{5}{6}$ , $|r| < 1$ , so its sum is $\frac{1/6}{1 - 5/6} = 1$ . The probabilities of all the outcomes add up to $1$ : the probability that no six ever appears is $0$ .

b) $E(N) = \sum_{n=1}^\infty n \cdot \frac{1}{6}\left(\frac{5}{6}\right)^{n-1} = \frac{1}{6}\sum_{n=1}^\infty n x^{n-1}$ with $x = \frac{5}{6}$ . The factor $\frac{1}{6}$ comes out of the sum because it does not depend on $n$ , and the power $n - 1$ matches $n x^{n-1}$ exactly, with no extra factor $x$ . By Exercise 5 a), $\sum_{n\ge1} n x^{n-1} = \frac{1}{(1 - x)^2}$ on $(-1, 1)$ , and $\frac{5}{6}$ lies in that OPEN interval, so the identity may be evaluated there: $E(N) = \frac{1}{6} \cdot \frac{1}{(1/6)^2} = \frac{1}{6} \cdot 36 = 6$ . On average, six rolls.

c) Exercise 5 c) gives $\sum_{n\ge1} n^2 x^{n-1} = \frac{1 + x}{(1 - x)^3}$ on $(-1, 1)$ . At $x = \frac{5}{6}$ : $\frac{11/6}{(1/6)^3} = \frac{11}{6} \cdot 216 = 396$ . So $E(N^2) = \frac{1}{6} \cdot 396 = 66$ and $\text{Var}(N) = 66 - 6^2 = 30$ , a standard deviation of $\sqrt{30}$ , about $5.5$ rolls since $5.5^2 = 30.25$ . The waiting time is very spread out, which the long tail of the figure already shows: a player who needs twelve rolls is not unlucky in any remarkable way.

d) A wins exactly when the first six comes on an odd-numbered roll: $P(\text{A wins}) = \sum_{k=0}^\infty P(N = 2k + 1) = \sum_{k=0}^\infty \frac{1}{6}\left(\frac{5}{6}\right)^{2k} = \frac{1}{6}\sum_{k=0}^\infty \left(\frac{25}{36}\right)^k = \frac{1}{6} \cdot \frac{1}{1 - 25/36} = \frac{1}{6} \cdot \frac{36}{11} = \frac{6}{11}$ . B wins with probability $\frac{5}{11}$ : moving first is worth a little under ten percentage points. Keeping only the odd terms of the series amounts to replacing $x$ by $x^2$ in the geometric series: the ratio becomes $\left(\frac{5}{6}\right)^2$ , still below $1$ .

e) The same computation with $x = 1 - p$ : $E(N) = p\sum_{n\ge1} n(1 - p)^{n-1} = p \cdot \frac{1}{(1 - (1 - p))^2} = \frac{p}{p^2} = \frac{1}{p}$ . It is valid because $0 < p \le 1$ puts $x = 1 - p$ in $[0, 1)$ , inside the open interval $(-1, 1)$ . For $p = 0$ , $x = 1$ is exactly the endpoint that Exercise 5 excluded: $\sum n \cdot 1^{n-1}$ diverges, and the formula $\frac{1}{p}$ breaks down with it; an event that never happens has no finite expected waiting time. For a double six, $p = \frac{1}{36}$ and $E(N) = 36$ throws of the pair of dice.

Exercise 10: A final exam problem: one series, its closed form and four numerical sums

Let $f(x) = \sum_{n=1}^\infty \frac{x^n}{n(n + 1)}$ . This question, typical of the last page of a MATH 141 final, links the interval of convergence, the series of $\ln(1 - x)$ and the sums of numerical series. Abel's theorem, stated in Exercise 6, may be used. Take $\ln 2 \approx 0.69$ and $\ln 3 \approx 1.10$ for orders of magnitude only.

a) Find the interval of convergence of the series.
b) Using $\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$ and the series of $\ln(1 - x)$ , show that $f(x) = 1 + \frac{1 - x}{x}\ln(1 - x)$ for $0 < |x| < 1$ . What is $f(0)$ , and is it consistent with the formula?
c) Deduce the exact value of $\sum_{n=1}^\infty \frac{1}{n(n + 1)2^n}$ , and check it against its first four terms.
d) Compute $f(1)$ by telescoping and compare with the limit of the formula of b) as $x \to 1^-$ . Deduce the exact value of $\sum_{n=1}^\infty \frac{(-1)^n}{n(n + 1)}$ .
e) Compute $\sum_{n=1}^\infty \frac{(-1)^n}{n(n + 1)2^n}$ exactly.

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Answers

a) $R = 1$ , interval $[-1, 1]$
b) $f(x) = 1 + \frac{1 - x}{x}\ln(1 - x)$ ; $f(0) = 0$ , the limit of the formula at $0$ .
c) $1 - \ln 2$
d) $f(1) = 1$ , equal to the limit; $\sum \frac{(-1)^n}{n(n + 1)} = 1 - 2\ln 2$
e) $1 - 3\ln\frac{3}{2}$

a) Ratio test: $\frac{n(n + 1)}{(n + 1)(n + 2)}|x| = \frac{n}{n + 2}|x| \to |x|$ , so $R = 1$ . At $x = \pm 1$ : $\left|\frac{(\pm 1)^n}{n(n + 1)}\right| = \frac{1}{n(n + 1)} \le \frac{1}{n^2}$ , convergent by comparison with $p = 2$ : absolute convergence at both endpoints. Interval $[-1, 1]$ .

b) Split: $f(x) = \sum_{n\ge1} \frac{x^n}{n} - \sum_{n\ge1} \frac{x^n}{n + 1}$ , legitimate for $|x| < 1$ because BOTH series converge there (splitting a convergent series into two divergent ones is the classic trap). The first is $-\ln(1 - x)$ . For the second, when $x \ne 0$ : $\sum_{n\ge1} \frac{x^n}{n + 1} = \frac{1}{x}\sum_{n\ge1} \frac{x^{n+1}}{n + 1} = \frac{1}{x}\sum_{k\ge2} \frac{x^k}{k} = \frac{1}{x}\left(-\ln(1 - x) - x\right)$ , the $k = 1$ term $x$ being missing. Hence $f(x) = -\ln(1 - x) + \frac{\ln(1 - x)}{x} + 1 = 1 + \frac{1 - x}{x}\ln(1 - x)$ . At $x = 0$ every term of the series contains $x^n$ with $n \ge 1$ , so $f(0) = 0$ . The formula is undefined at $0$ , but its limit is $1 + \lim_{x\to0}(1 - x)\frac{\ln(1 - x)}{x} = 1 + 1 \cdot (-1) = 0$ , by L'Hôpital's rule on $\frac{\ln(1 - x)}{x}$ . Consistent, as it must be: a power series is continuous on its open interval. The solution figure draws $f$ on $[-1, 1]$ .

c) $x = \frac{1}{2}$ is inside the interval: $\sum \frac{1}{n(n + 1)2^n} = f\left(\frac{1}{2}\right) = 1 + \frac{1/2}{1/2}\ln\frac{1}{2} = 1 - \ln 2$ , about $1 - 0.69 = 0.31$ . Check: $\frac{1}{4} + \frac{1}{24} + \frac{1}{96} + \frac{1}{320} \approx 0.250 + 0.042 + 0.010 + 0.003 = 0.305$ , and the terms left are positive and small, the next being $\frac{1}{960}$ . Agreement.

d) At $x = 1$ : $f(1) = \sum \frac{1}{n(n + 1)} = \sum\left(\frac{1}{n} - \frac{1}{n + 1}\right)$ , telescoping, $s_N = 1 - \frac{1}{N + 1} \to 1$ . The formula: with $t = 1 - x \to 0^+$ , $\frac{1 - x}{x}\ln(1 - x) = \frac{t\ln t}{1 - t}$ and $t\ln t \to 0$ (L'Hôpital's rule on $\frac{\ln t}{1/t}$ ), so the limit is $1$ . The two agree, as Abel's theorem promises. At the other endpoint, $x = -1$ , the series converges and the formula is continuous, so Abel's theorem gives $\sum \frac{(-1)^n}{n(n + 1)} = f(-1) = 1 + \frac{2}{-1}\ln 2 = 1 - 2\ln 2$ , about $1 - 1.38 = -0.38$ . The partial sums $-\frac{1}{2}$ , $-\frac{1}{3}$ , $-\frac{5}{12}$ of this alternating series bracket that value, as they must.

e) $x = -\frac{1}{2}$ is inside: $f\left(-\frac{1}{2}\right) = 1 + \frac{3/2}{-1/2}\ln\frac{3}{2} = 1 - 3\ln\frac{3}{2}$ . With $\ln\frac{3}{2} = \ln 3 - \ln 2 \approx 1.10 - 0.69 = 0.41$ , this is about $1 - 1.23 = -0.23$ , and the first terms $-\frac{1}{4} + \frac{1}{24} - \frac{1}{96} + \frac{1}{320} \approx -0.216$ agree to the precision of the logarithms used. The factor $\frac{1 - x}{x}$ is where the sign slips happen: at $x = -\frac{1}{2}$ , $1 - x = \frac{3}{2}$ , not $\frac{1}{2}$ , and $x$ is negative.

Corrected exercises: power series (MATH 141)

Course recap

Part A: the basics (/50)

Exercise 1: Radius first, then each endpoint with its own test

Exercise 2: Centre, gaps and extreme radii: read the series before testing it

Exercise 3: Building series from 1/(1 - u): substitute, factor, recentre

Exercise 4: Partial sums against the function: inside, at and beyond the radius

Exercise 5: Differentiate term by term, then evaluate: 1/(1 - x) squared and its cousins

Part B: problems and reasoning (/50)

Exercise 6: Integrate term by term: ln(1 + x), arctan x and the constant

Exercise 7: The same radius, not the same interval: endpoints lost and gained

Exercise 8: Five statements to correct

Exercise 9: Waiting for a six: expected values as power series

Exercise 10: A final exam problem: one series, its closed form and four numerical sums

See also

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