Corrected exercises: power series (MATH 141)

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: $[-3, 3)$.

$a_n = \frac{n\,x^n}{4^n}$: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{n + 1}{n} \cdot \frac{|x|}{4} \to \frac{|x|}{4}$, 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.

$a_n = \frac{(-1)^n x^n}{\sqrt{n + 1}}$, $n \ge 0$: $\left|\frac{a_{n+1}}{a_n}\right| = |x|\sqrt{\frac{n + 1}{n + 2}} \to |x|$, 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.

$a_n = \frac{x^n}{n^2\,5^n}$: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{5} \cdot \frac{n^2}{(n + 1)^2} \to \frac{|x|}{5}$, $R = 5$. 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: $[-5, 5]$.

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.

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.

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 $|r| = 1$.

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n + 1)!\,|x - 1|^{n+1}}{n!\,|x - 1|^n} = (n + 1)|x - 1|$. 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.

$\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x + 2|^{n+1}}{(n + 1)!} \cdot \frac{n!}{|x + 2|^n} = \frac{|x + 2|}{n + 1} \to 0 < 1$ 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.

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.

$\frac{1}{1 + 3x} = \frac{1}{1 - (-3x)}$, 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 $\frac{1}{1 + 0} = 1$.

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 $16$.

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 $\frac{1}{7} = \frac{1}{2 + 5}$.

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 $(-1, 1)$.

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 $1$.

$\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \sum_{n=0}^\infty (-x^2)^n = \sum_{n=0}^\infty (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots$, 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.

$S_N(x)$ 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.

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 $0$.

$S_N(1) = \frac{1 - (-1)^{N+1}}{2}$ 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.

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.