Corrected exercises: geometric series, telescoping series and the divergence test (MATH 141)

Add the terms one at a time: $s_1 = \frac{2}{3}$, $s_2 = \frac{2}{3} + \frac{2}{9} = \frac{8}{9}$, $s_3 = \frac{8}{9} + \frac{2}{27} = \frac{26}{27}$, $s_4 = \frac{26}{27} + \frac{2}{81} = \frac{80}{81}$. A pattern is already visible: each partial sum is $1$ minus a power of $\frac{1}{3}$, namely $\frac{3^n - 1}{3^n}$. Computing the first three or four partial sums is never wasted on an exam: it is the only way to see what the series does before proving it, and it checks every closed form you will write afterwards.

$s_n = \frac{2}{3} + \frac{2}{3^2} + \dots + \frac{2}{3^n}$ and $\frac{1}{3}s_n = \frac{2}{3^2} + \dots + \frac{2}{3^n} + \frac{2}{3^{n+1}}$. Subtracting, every term cancels except the first of $s_n$ and the last of $\frac{1}{3}s_n$: $\frac{2}{3}s_n = \frac{2}{3} - \frac{2}{3^{n+1}}$. Multiply by $\frac{3}{2}$: $s_n = 1 - \frac{1}{3^n}$. Check with a): $n = 3$ gives $1 - \frac{1}{27} = \frac{26}{27}$. This subtraction is exactly the proof of the geometric sum formula, and it is worth knowing because it works on a FINITE sum, where nothing can go wrong.

Since $\frac{1}{3^n} \to 0$, $s_n \to 1$: the series converges and $\sum_{n=1}^{\infty} \frac{2}{3^n} = 1$. On the figure the points climb (every term is positive, so $s_n$ increases) and flatten out just below height $1$, which they never reach: $s_n < 1$ for every $n$. The sum is the LEVEL the partial sums approach, not a partial sum. Writing the conclusion as a limit of $s_n$ is the sentence the marker looks for: the series converges BECAUSE its partial sums have a limit.

The first partial sum is the first term: $a_1 = s_1 = \frac{4}{2} = 2$. For $n \ge 2$, $a_n = s_n - s_{n-1} = \frac{4n}{n+1} - \frac{4(n-1)}{n} = \frac{4n^2 - 4(n-1)(n+1)}{n(n+1)} = \frac{4n^2 - 4n^2 + 4}{n(n+1)} = \frac{4}{n(n+1)}$. The formula also gives $a_1 = \frac{4}{2} = 2$, so it holds for all $n \ge 1$. The sum is $\lim s_n = \lim \frac{4n}{n+1} = \lim \frac{4}{1 + 1/n} = 4$. Note the direction: from $s_n$ back to $a_n$ is one subtraction, and the sum is read on $s_n$ directly, with no formula at all.

He has taken the limit of the wrong sequence. $\lim a_n = \lim \frac{4}{n(n+1)} = 0$ says only that the terms become small; the sum is $\lim s_n = 4$. Two sanity checks kill the answer $0$ at once. First, every term is positive, so $s_n \ge s_1 = 2$ for all $n$, and a limit of numbers all at least $2$ is at least $2$. Second, if a series converges, its terms ALWAYS tend to $0$, so $\lim a_n = 0$ would be the sum of every convergent series, which is absurd. Two sequences live in every series question, $(a_n)$ and $(s_n)$: name the one you are taking the limit of.

Rewrite the term so that the exponents match: $\frac{4^n}{5^{n-1}} = 4 \cdot \frac{4^{n-1}}{5^{n-1}} = 4\left(\frac{4}{5}\right)^{n-1}$. First term ($n = 1$): $\frac{4}{1} = 4$. Ratio: $\frac{a_{n+1}}{a_n} = \frac{4^{n+1}}{5^n} \cdot \frac{5^{n-1}}{4^n} = \frac{4}{5}$. Since $|r| = \frac{4}{5} < 1$, the series converges and its sum is $\frac{4}{1 - 4/5} = \frac{4}{1/5} = 20$. The trap is to read $r = \frac{4}{5}$ correctly but take the first term to be $1$ (as if the series were $\sum (4/5)^n$ from $n = 0$), giving $5$, a quarter of the answer.

The series starts at $n = 2$, so the first term is $a_2 = \frac{(-3)^3}{4^2} = -\frac{27}{16}$. The ratio is $\frac{(-3)^{n+2}}{4^{n+1}} \cdot \frac{4^n}{(-3)^{n+1}} = -\frac{3}{4}$. Since $|r| = \frac{3}{4} < 1$, it converges, to $\frac{-27/16}{1 - (-3/4)} = \frac{-27/16}{7/4} = -\frac{27}{16} \cdot \frac{4}{7} = -\frac{27}{28}$. Two signs to keep: the ratio is NEGATIVE, so $1 - r = \frac{7}{4}$, not $\frac{1}{4}$; and the first term is negative because $n + 1 = 3$ is odd. Quick check: $a_2 + a_3 = -\frac{27}{16} + \frac{81}{64} = -\frac{27}{64}$, and the sum $-\frac{27}{28}$ lies between $a_2$ and $a_2 + a_3$, as it must for a series whose partial sums alternate around their limit.

$2^{2n+1} = 2 \cdot 4^n$, so the term is $2\left(\frac{4}{3}\right)^n$: the exponent $2n$ hides a ratio of $4$, not $2$. First term ($n = 0$): $2$. Ratio: $\frac{4}{3}$. Since $|r| = \frac{4}{3} \ge 1$, the series diverges. The divergence test says the same thing independently: $2\left(\frac{4}{3}\right)^n \to \infty$, so the terms do not tend to $0$. Plugging into $\frac{a}{1 - r}$ anyway gives $\frac{2}{1 - 4/3} = -6$, a negative number for a sum of positive terms: the formula is valid ONLY when $|r| < 1$, and it does not warn you when you misuse it.

$\frac{2^n + 3^n}{6^n} = \left(\frac{1}{3}\right)^n + \left(\frac{1}{2}\right)^n$. Both $\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$ and $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ are geometric with $|r| < 1$, hence convergent, and the sum of two CONVERGENT series may be computed term by term: $\sum (a_n + b_n) = \sum a_n + \sum b_n$. So the sum is $\frac{1/3}{1 - 1/3} + \frac{1/2}{1 - 1/2} = \frac{1}{2} + 1 = \frac{3}{2}$. The justification is not decoration: splitting is legal because each piece converges, and it would be illegal if either one diverged (Exercise 7 shows what goes wrong then).

$e^{-2n} = \left(e^{-2}\right)^n$: first term $e^{-2}$, ratio $e^{-2} = \frac{1}{e^2} < 1$ since $e^2 > 1$. The sum is $\frac{e^{-2}}{1 - e^{-2}}$; multiplying top and bottom by $e^2$ gives $\frac{1}{e^2 - 1}$, the exact answer expected without a calculator. For the second series, $\frac{\pi^n}{3^{n+1}} = \frac{1}{3}\left(\frac{\pi}{3}\right)^n$, ratio $\frac{\pi}{3}$. Deciding $|r| < 1$ or not needs no machine: $\pi > 3$, so $\frac{\pi}{3} > 1$ and the series diverges. It looks harmless because $\frac{\pi}{3}$ is close to $1$, but a ratio just above $1$ still makes the terms grow: $\left(\frac{\pi}{3}\right)^n \to \infty$.

Partial fractions: $\frac{1}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}$, so $1 = A(n+2) + Bn$; $n = 0$ gives $A = \frac{1}{2}$ and $n = -2$ gives $B = -\frac{1}{2}$. Thus $a_n = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+2}\right)$. Now write $s_N$ with enough terms: $s_N = \frac{1}{2}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{N-1} - \frac{1}{N+1}\right) + \left(\frac{1}{N} - \frac{1}{N+2}\right)\right]$. The gap is TWO, so each negative term cancels with a positive term two brackets later, and two terms survive at each end: $s_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right)$. As $N \to \infty$, $s_N \to \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$. Keeping only the $1$ (as for a gap of one) gives $\frac{1}{2}$: writing three or four brackets is what shows how many terms survive.

$\ln\frac{n}{n+1} = \ln n - \ln(n+1)$. Then $s_N = (\ln 1 - \ln 2) + (\ln 2 - \ln 3) + \dots + (\ln N - \ln(N+1)) = \ln 1 - \ln(N+1) = -\ln(N+1)$. Since $\ln(N+1) \to \infty$, $s_N \to -\infty$ and the series DIVERGES. Yet its terms tend to $\ln 1 = 0$. Here is the chapter's central warning in its purest form: the cancellation is perfect, the terms vanish, and the sum is still infinite, because what survives in $s_N$ is a quantity that does not have a finite limit.

$s_N = (\arctan 2 - \arctan 1) + (\arctan 3 - \arctan 2) + \dots + (\arctan(N+1) - \arctan N) = \arctan(N+1) - \arctan 1$. With $\arctan 1 = \frac{\pi}{4}$ and $\arctan(N+1) \to \frac{\pi}{2}$, the series converges to $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. Compare with b): same structure $b_{n+1} - b_n$, opposite verdicts. The telescoping series $\sum (b_{n+1} - b_n)$ converges exactly when $(b_n)$ has a finite limit $L$, and then its sum is $L - b_1$. Here $b_n = \arctan n$ is bounded by $\frac{\pi}{2}$; in b), $b_n = -\ln n$ is not.

$(n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$, so $\frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}$. This is the partial fraction decomposition obtained by recognition rather than by solving for constants; a full decomposition with $\frac{A}{n} + \frac{B}{n^2} + \frac{C}{n+1} + \frac{D}{(n+1)^2}$ gives the same thing with $A = C = 0$, at four times the cost. Then $s_N = 1 - \frac{1}{(N+1)^2} \to 1$: the series converges to $1$. Check: $a_1 = \frac{3}{4}$ and $s_1 = 1 - \frac{1}{4} = \frac{3}{4}$.

The remainder is exact here, not an estimate: $\frac{3}{4} - s_N = \frac{1}{2}\left(\frac{1}{N+1} + \frac{1}{N+2}\right)$. We need $\frac{1}{N+1} + \frac{1}{N+2} < \frac{1}{50}$. For $N = 99$: $\frac{1}{100} + \frac{1}{101} < \frac{1}{100} + \frac{1}{100} = \frac{1}{50}$, which works. For $N = 98$: $\frac{1}{99} + \frac{1}{100} > \frac{1}{100} + \frac{1}{100} = \frac{1}{50}$, which fails. And the remainder decreases with $N$, so $N = 99$ is the smallest. This is the privilege of a telescoping series: since $s_N$ is known in closed form, the error is known exactly, and a question of precision becomes an inequality solved by hand.

Divide numerator and denominator by $n^2$: $a_n = \frac{2 - 1/n^2}{5 + 1/n} \to \frac{2}{5}$. Since $\frac{2}{5} \ne 0$, the series diverges by the divergence test. The picture behind it: from some point on, every term is close to $0.4$, so each new term adds about $0.4$ to the partial sum, and $s_n$ grows roughly like $0.4n$. A series cannot converge while adding a fixed amount at every step.

Set $x = \frac{1}{n}$: $n\sin\frac{1}{n} = \frac{\sin x}{x}$ with $x \to 0^+$, and $\lim_{x\to 0} \frac{\sin x}{x} = 1$. So $a_n \to 1 \ne 0$ and the series diverges. The trap is to see $\sin\frac{1}{n} \to 0$ and stop there: the factor $n$ in front grows at exactly the rate needed to compensate. Always take the limit of the WHOLE term $a_n$, never of a piece of it.

The limit $\left(1 + \frac{x}{n}\right)^n \to e^x$ with $x = -2$ gives $a_n \to e^{-2}$. To justify it, take the logarithm: $n\ln\left(1 - \frac{2}{n}\right) = \frac{\ln(1 - 2t)}{t}$ with $t = \frac{1}{n} \to 0^+$, a $\frac{0}{0}$ form; L'Hôpital's rule on the function of $t$ gives $\frac{-2/(1 - 2t)}{1} \to -2$, so $a_n \to e^{-2}$. Since $e^{-2} \ne 0$ (an exponential is never $0$), the series diverges. The classic slip: $1 - \frac{2}{n} \to 1$, so $a_n \to 1^n = 1$. The exponent grows with $n$, so the limit of the base cannot be substituted; $1^\infty$ is an indeterminate form. That slip happens to reach the right verdict here, but with a wrong limit, and the limit is what is marked.

$\cos(n\pi) = (-1)^n$: the terms are $-1, 1, -1, 1, \dots$ and $\lim a_n$ does not exist. The test covers this case too: its hypothesis is that $a_n$ does not tend to $0$, which includes having no limit at all. So the series diverges. Its partial sums are $-1, 0, -1, 0, \dots$, which confirms it: they have no limit either.

For both series $\lim a_n = 0$, so the divergence test says NOTHING, and both verdicts must come from the partial sums. For $\sum \frac{1}{\sqrt{n}}$: each of the $n$ terms of $s_n$ is at least the smallest one, $\frac{1}{\sqrt{n}}$, so $s_n \ge n \cdot \frac{1}{\sqrt{n}} = \sqrt{n} \to \infty$: the series diverges. For $\sum \frac{1}{n^2 + n} = \sum \left(\frac{1}{n} - \frac{1}{n+1}\right)$: telescoping, $s_n = 1 - \frac{1}{n+1} \to 1$, the series converges to $1$. Same verdict from the test (silent), opposite outcomes: this is the whole meaning of the sentence the converse of the divergence test is false. Chapter 19 will give general tests for such series; here the partial sums were enough.

$s_1 = 1$, $s_2 = \frac{3}{2}$, $s_4 = \frac{3}{2} + \frac{1}{3} + \frac{1}{4} = \frac{18 + 4 + 3}{12} = \frac{25}{12}$, $s_8 = \frac{25}{12} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$. Over the common denominator $840$: $\frac{1750 + 168 + 140 + 120 + 105}{840} = \frac{2283}{840} = \frac{761}{280}$, a little under $2.72$. Compare with b): $1 + \frac{0}{2} = 1$, $1 + \frac{1}{2}$, $1 + \frac{2}{2} = 2 < \frac{25}{12}$ and $1 + \frac{3}{2} = \frac{5}{2} < \frac{761}{280}$. The bound holds and is not far off, which the figure shows: the open dots sit just under the filled ones.

The $j$th block runs from $\frac{1}{2^{j-1}+1}$ to $\frac{1}{2^j}$: it holds $2^j - 2^{j-1} = 2^{j-1}$ terms, and the smallest of them is the last one, $\frac{1}{2^j}$. So the block is at least $2^{j-1} \cdot \frac{1}{2^j} = \frac{1}{2}$. For instance $\frac{1}{3} + \frac{1}{4} \ge \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ and $\frac{1}{5} + \dots + \frac{1}{8} \ge 4 \cdot \frac{1}{8} = \frac{1}{2}$. The sum $s_{2^k}$ is $1$, plus $\frac{1}{2}$, plus $k - 1$ blocks ($j = 2$ to $k$), so $s_{2^k} \ge 1 + \frac{1}{2} + \frac{k-1}{2} = 1 + \frac{k}{2}$. The gesture is to bound a whole block by its number of terms times its smallest term, which is the same gesture as $s_n \ge \sqrt{n}$ in Exercise 4.

Every term is positive, so $(s_n)$ is increasing. By b), $s_{2^k} \ge 1 + \frac{k}{2}$ exceeds any given number $M$ once $k > 2M$. An increasing sequence that exceeds every $M$ at some index stays above $M$ afterwards, so $s_n \to \infty$ and the harmonic series diverges. Controlling $s_n$ at $n = 2^k$ only is enough precisely because $s_n$ is increasing: between two powers of $2$ it cannot fall back. Without that monotonicity, a subsequence tending to infinity would still prove divergence, but not divergence to $\infty$.

Now group from $\frac{1}{2^j}$ to $\frac{1}{2^{j+1}-1}$: $2^j$ terms, the LARGEST being $\frac{1}{2^j}$, so each block is at most $2^j \cdot \frac{1}{2^j} = 1$. The blocks $j = 0, 1, \dots, k-1$ make up exactly $s_{2^k - 1}$, so $s_{2^k - 1} \le k$. With $k = 10$: $s_{1023} \le 10$, and since $(s_n)$ increases, $s_n \le 10$ for every $n \le 1023$. From b) with $k = 18$: $s_{2^{18}} \ge 1 + 9 = 10$, and the inequality is strict because the blocks from $j = 2$ on are strictly larger than $\frac{1}{2}$ (their terms are not all equal). So $n = 2^{18} = 262\,144$ guarantees $s_n > 10$. The true first $n$ lies between these two bounds (it is $12\,367$), and the two groupings squeeze $s_{2^k}$ between $1 + \frac{k}{2}$ and $k + 1$.

A small increment does not mean a finite total. From $n = 1000$ on each term is below $0.001$, but there are infinitely many of them, and d) shows that the sum still passes $10$ (between $n = 1024$ and $n = 262\,144$), then $100$, then any number; it simply takes astronomically many terms, since doubling $n$ adds only about $\frac{1}{2}$ or more. A computer adding terms can suggest a convergence that does not happen, and it can never prove one. The verdict comes from an argument on $s_n$, as in b) and c), never from a table of values: this is exactly why the harmonic series is the example every exam uses to test the false converse of the divergence test.