Corrected exercises: alternating series, absolute and conditional convergence (MATH 141)

Strip the sign: $a_n = (-1)^{n-1} b_n$ with $b_n = \frac{1}{2n-1}$. Hypothesis 1, positive: $2n - 1 \ge 1$, so $b_n > 0$. Hypothesis 2, decreasing: $2(n+1) - 1 = 2n + 1 > 2n - 1$, and a larger positive denominator gives a smaller fraction, so $b_{n+1} = \frac{1}{2n+1} < \frac{1}{2n-1} = b_n$. Hypothesis 3, limit: $\lim_{n \to \infty} \frac{1}{2n-1} = 0$. All three hold, so by the Alternating Series Test the series converges. The marker gives one line per hypothesis; writing only $b_n \to 0$ and concluding is worth half the marks at most, because the limit alone does not make an alternating series converge (Exercise 7 builds one that diverges).

Here $b_n = \frac{n}{n+1} = \frac{1}{1 + 1/n}$, which tends to $1$, not to $0$. The third hypothesis fails, so the Alternating Series Test says NOTHING: it is a one-way test, and a failed hypothesis is not a verdict. The verdict comes from the Divergence Test of chapter 18: the terms $a_n = (-1)^n \frac{n}{n+1}$ approach $1$ along even $n$ and $-1$ along odd $n$, so $\lim a_n$ does not exist, in particular it is not $0$, and the series diverges. Writing the series diverges because the alternating series test fails loses the mark for the reason even though the conclusion is right.

For every integer $n$, $\cos(n\pi) = (-1)^n$: $\cos \pi = -1$, $\cos 2\pi = 1$, and so on. The series is $\sum \frac{(-1)^n}{\sqrt{n}}$, alternating, with $b_n = \frac{1}{\sqrt{n}} > 0$. Decreasing: $\sqrt{n+1} > \sqrt{n}$, so $\frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}}$. Limit: $\frac{1}{\sqrt{n}} \to 0$. The series converges. It starts with a negative term, $-1$, and that changes nothing: the test applies to $\sum (-1)^n b_n$ exactly as to $\sum (-1)^{n-1} b_n$, the two differing by a factor $-1$.

It does not matter: removing or changing finitely many terms never changes whether a series converges, only its sum. So apply the test to $\sum_{n=2}^{\infty} (-1)^n \sin\frac{\pi}{n}$. For $n \ge 2$, $0 < \frac{\pi}{n} \le \frac{\pi}{2}$, where $\sin$ is positive, so $b_n > 0$. Decreasing: $0 < \frac{\pi}{n+1} < \frac{\pi}{n} \le \frac{\pi}{2}$ and $\sin$ is increasing on $\left[0, \frac{\pi}{2}\right]$, so $b_{n+1} < b_n$. Limit: $\frac{\pi}{n} \to 0$ and $\sin$ is continuous at $0$, so $b_n \to \sin 0 = 0$. The series converges. Note the argument for the decrease: it uses where the angle lives. For $n = 1$ the angle is $\pi$, outside $\left[0, \frac{\pi}{2}\right]$, which is precisely why the first term had to be set aside.

The first term is $+1$, so the odd partial sums $S_1, S_3, S_5, \ldots$ (orange) end on an added term and lie ABOVE $\frac{\pi}{4}$, the even ones $S_2, S_4, \ldots$ (blue) end on a subtracted term and lie BELOW. The sum is squeezed between any two consecutive partial sums. $S_{10}$ is even, so it is below, and the next step adds $b_{11} = \frac{1}{21}$ and jumps over the limit: $0 < \frac{\pi}{4} - S_{10} < \frac{1}{21}$. The picture explains the test: each step is shorter than the previous one (decreasing) and the steps shrink to nothing (limit $0$), so the zigzag closes on a single number.

$b_2 = \frac{\ln 2}{2}$ and $b_3 = \frac{\ln 3}{3}$. Multiply both by $6$: the comparison is between $3\ln 2 = \ln 8$ and $2\ln 3 = \ln 9$, and $\ln$ is increasing, so $\ln 8 < \ln 9$ and $b_2 < b_3$. Also $b_1 = \frac{\ln 1}{1} = 0 < b_2$. The sequence RISES at first: it is not decreasing from $n = 1$, and a proof that begins it is clear that $b_n$ decreases is proving something false. The test only needs the decrease from some rank on, so the question becomes: from which rank?

Quotient rule: $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$. The denominator is positive, so $f'(x) < 0$ exactly when $\ln x > 1$, that is $x > e$. Since $2 < e < 3$, $f$ is decreasing on $[3, \infty)$ and $b_{n+1} < b_n$ for every $n \ge 3$; the figure shows the peak at $x = e$. Limit: $\lim_{x \to \infty} \frac{\ln x}{x}$ has the form $\frac{\infty}{\infty}$, and L'Hôpital's rule on the FUNCTION gives $\lim \frac{1/x}{1} = 0$; since $b_n = f(n)$, $\lim b_n = 0$. From $n = 3$ on, $b_n > 0$, decreasing, limit $0$: the series converges by the Alternating Series Test, the first two terms changing only the value of the sum. Writing $\frac{d}{dn}$ of $b_n$ or L'Hôpital on $b_n$ is the chapter 17 trap: a sequence has no derivative, its function does.

$g'(x) = \frac{\frac{1}{2\sqrt{x}}(x+4) - \sqrt{x}}{(x+4)^2} = \frac{(x + 4) - 2x}{2\sqrt{x}(x+4)^2} = \frac{4 - x}{2\sqrt{x}(x+4)^2}$, after multiplying top and bottom by $2\sqrt{x}$. The denominator is positive for $x > 0$, so $g'(x) < 0$ for $x > 4$: $b_n$ decreases from $n = 4$ on. Limit: divide top and bottom by $n$, $b_n = \frac{1/\sqrt{n}}{1 + 4/n} \to \frac{0}{1} = 0$. And $b_n > 0$. The series converges by the Alternating Series Test applied from $n = 4$.

The denominator $n + 4$ grows, but so does the numerator $\sqrt{n}$, and a quotient of two growing quantities can go either way: this one grows until $n = 4$. The derivative is what settles the race. By hand: $b_3 < b_4$ reads $\frac{\sqrt{3}}{7} < \frac{2}{8} = \frac{1}{4}$, that is $4\sqrt{3} < 7$, and squaring two positive numbers keeps the order: $48 < 49$, true. $b_4 > b_5$ reads $\frac{1}{4} > \frac{\sqrt{5}}{9}$, that is $9 > 4\sqrt{5}$, and squaring: $81 > 80$, true. The margins are tiny, which is exactly why intuition fails here and a sign study is required.

Absolute convergence asks about $\sum b_n$, the series of sizes, and it is a chapter 19 question. For a): for $n \ge 3$, $\ln n > 1$, so $\frac{\ln n}{n} > \frac{1}{n}$; the harmonic series diverges, so $\sum \frac{\ln n}{n}$ diverges by direct comparison (the inequality points the right way: bigger than a divergent series). For c): $\frac{\sqrt{n}/(n+4)}{1/\sqrt{n}} = \frac{n}{n+4} \to 1$, a finite positive limit, and $\sum \frac{1}{\sqrt{n}}$ is a divergent $p$-series ($p = \frac{1}{2} \le 1$), so $\sum b_n$ diverges by limit comparison. Both series converge, but not absolutely: they converge CONDITIONALLY, thanks to the cancellation between signs only.

$b_n = \frac{1}{n^3}$ is positive; $(n+1)^3 > n^3$ gives $b_{n+1} < b_n$; and $\frac{1}{n^3} \to 0$. The test applies, the series converges, and the estimation theorem may be used. The theorem has exactly the same hypotheses as the test: quoting the bound on a series whose $b_n$ is not decreasing is not allowed (Exercise 8 c).

$S_4 = 1 - \frac{1}{8} + \frac{1}{27} - \frac{1}{64}$. The common denominator is $\text{lcm}(8, 27, 64) = 64 \cdot 27 = 1728$: $S_4 = \frac{1728 - 216 + 64 - 27}{1728} = \frac{1549}{1728}$. The first term left out is $+\frac{1}{5^3}$, so $|S - S_4| \le b_5 = \frac{1}{125} = 0.008$. The bound uses $b_5$, not $b_4 = \frac{1}{64}$: the bound with $b_4$ is true but wasteful, and the question asks for the theorem's bound.

The first omitted term, $+\frac{1}{125}$, is positive, so the error $S - S_4$ is positive: $S_4$ is an UNDERESTIMATE. Precisely, $S$ lies between $S_4$ and $S_5 = S_4 + \frac{1}{125}$: $\frac{1549}{1728} < S < \frac{1549}{1728} + \frac{1}{125}$. With the common denominator $216000 = 1728 \times 125$: $S_5 = \frac{1549 \cdot 125 + 1728}{216000} = \frac{193625 + 1728}{216000} = \frac{195353}{216000}$. The interval $(S_4, S_5)$ has length $b_5 = \frac{1}{125}$, and the figure of the solution places it on the number line: $S_1$, $S_3$, $S_5$ on the right, $S_2$, $S_4$ on the left, closing in.

$0.895 \times 1728 = 1546.56 < 1549$, so $S_4 = \frac{1549}{1728} > 0.895$. And $0.905 \times 216000 = 195480 > 195353$, so $S_5 = \frac{195353}{216000} < 0.905$. Hence $0.895 < S < 0.905$, and every number of that interval rounds to $0.90$. This is how a no-calculator exam asks for decimals: the bracket is exact, the rounding follows from two multiplications. (The true value is about $0.9015$, well inside.)

$S_{2m} = (b_1 - b_2) + (b_3 - b_4) + \cdots + (b_{2m-1} - b_{2m})$. Each bracket is $\ge 0$ because $(b_n)$ is DECREASING, so going from $S_{2m}$ to $S_{2m+2}$ adds $b_{2m+1} - b_{2m+2} \ge 0$: the even sums increase. Grouped the other way, $S_{2m} = b_1 - (b_2 - b_3) - \cdots - (b_{2m-2} - b_{2m-1}) - b_{2m}$, where every subtracted piece is $\ge 0$, again by the decrease, so $S_{2m} \le b_1$. An increasing sequence bounded above converges (Monotone Sequence Theorem, chapter 17): $S_{2m} \to S$. Finally $S_{2m+1} = S_{2m} + b_{2m+1}$, and $b_{2m+1} \to 0$ by the LIMIT hypothesis, so the odd sums have the same limit $S$. The same picture gives the estimate: the even sums climb to $S$ and the odd sums descend to it, so $S$ is always between two consecutive sums, at distance less than the step $b_{n+1}$ between them.

$b_n = \frac{1}{n^2}$ is positive, decreasing, with limit $0$, so the theorem applies. We need $b_{n+1} = \frac{1}{(n+1)^2} \le \frac{1}{1000}$, that is $(n+1)^2 \ge 1000$. Now $31^2 = 961 < 1000$ and $32^2 = 1024 \ge 1000$, so the smallest possibility is $n + 1 = 32$, $n = 31$. The sum $S_{31}$ of the first $31$ terms is guaranteed within $0.001$ of $S$. Checking the two squares on either side of $1000$ is the proof that $31$ is the smallest $n$ the theorem allows; writing $n \ge \sqrt{1000} - 1$ and stopping there leaves the answer unfinished.

$b_n = \frac{1}{(2n)!} > 0$; decreasing because $(2n+2)! = (2n+2)(2n+1) \cdot (2n)! > (2n)!$; and $\frac{1}{(2n)!} \to 0$. The index starts at $n = 0$, so $S_N = \sum_{n=0}^{N}$ has $N + 1$ terms and the first omitted term is $b_{N+1} = \frac{1}{(2N+2)!}$. We need $(2N+2)! \ge 10^4$: $6! = 720$ is too small, $8! = 40320$ is enough, so $2N + 2 = 8$ and $N = 3$. The estimate is $S_3 = 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720} = \frac{720 - 360 + 30 - 1}{720} = \frac{389}{720}$, made of FOUR terms, with error at most $\frac{1}{8!} = \frac{1}{40320} < 10^{-4}$. The factorials are handled by the estimation theorem alone; no other test is needed to see that this series converges very fast.

$b_{n+1} = \frac{1}{\sqrt{n+1}} \le \frac{1}{100}$ means $\sqrt{n+1} \ge 100$, that is $n + 1 \ge 10000$: $n = 9999$ terms. Compare with a) and b): $31$ terms for three decimals with $\frac{1}{n^2}$, four terms for four decimals with factorials, ten thousand terms for two decimals with $\frac{1}{\sqrt{n}}$. The bound reads the SPEED at which $b_n$ goes to $0$: a conditionally convergent alternating series usually converges very slowly, and is useless for computing.

The theorem bounds the error by the first omitted term, $b_{n+1}$, so the second student is right and the answer is $31$. The first student solves $b_n \le 0.001$, gets $n = 32$ and sums one term too many: the result is still within $0.001$, but it is not the number the theorem gives, and the question asks for that number. The third student confuses the index with the count: when the sum starts at $n = 0$, $S_3$ contains $n = 0, 1, 2, 3$, four terms. Both slips come from the same source, not writing down the first omitted term explicitly before solving.

$S_3 = 1 - \frac{1}{4} + \frac{1}{9} = \frac{36 - 9 + 4}{36} = \frac{31}{36}$, and $S_4 = \frac{31}{36} - \frac{1}{16} = \frac{124 - 9}{144} = \frac{115}{144}$. The sum lies between two consecutive partial sums, the odd one above since the last term added was positive: $\frac{115}{144} < S < \frac{31}{36}$, an interval of length $b_4 = \frac{1}{16}$. (The value, $\frac{\pi^2}{12} \approx 0.82$, is indeed between $0.79$ and $0.87$.)

$|a_n| = \frac{1}{n^{3/2}}$, a $p$-series with $p = \frac{3}{2} > 1$: it converges. So the series converges absolutely, hence converges. No Alternating Series Test is needed, and running it would only prove convergence, which is the weaker statement: when asked to classify, answering converges by the alternating series test for a) loses the mark for absolutely.

$|a_n| = \frac{n}{n^2+1}$. Limit comparison with $\frac{1}{n}$: $\frac{n/(n^2+1)}{1/n} = \frac{n^2}{n^2+1} \to 1$, finite and positive, and $\sum \frac{1}{n}$ diverges, so $\sum |a_n|$ diverges: not absolutely convergent. Now the signed series, $b_n = \frac{n}{n^2+1} > 0$. Decreasing: $h(x) = \frac{x}{x^2+1}$ has $h'(x) = \frac{(x^2+1) - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2} < 0$ for $x > 1$, so $b_n$ decreases from $n = 1$ ($b_1 = \frac{1}{2} > b_2 = \frac{2}{5}$). Limit: $\frac{n}{n^2+1} = \frac{1/n}{1 + 1/n^2} \to 0$. By the Alternating Series Test the series converges: CONDITIONALLY.

The signs of $\sin n$ are $+, +, +, -, -, -, +, \ldots$, following the position of $n$ radians on the circle: this is NOT an alternating series and the Alternating Series Test cannot be used, not even to fail. Go to absolute values: $0 \le \frac{|\sin n|}{n^2} \le \frac{1}{n^2}$, and $\sum \frac{1}{n^2}$ converges ($p = 2$), so $\sum |a_n|$ converges by direct comparison. The series converges absolutely, hence converges. The figure of the solution shows the sizes $|a_n|$, blue for a positive term and red for a negative one, all under the curve $\frac{1}{n^2}$: absolute convergence is the only tool for irregular signs, and the reason the theorem absolute implies convergent is worth knowing.

For $n \ge 2$, $0 < \ln n < n$, so $\frac{1}{\ln n} > \frac{1}{n}$, and $\sum_{n \ge 2} \frac{1}{n}$ diverges: $\sum |a_n|$ diverges by direct comparison. Signed series: $b_n = \frac{1}{\ln n} > 0$ for $n \ge 2$, decreasing because $\ln$ is increasing, and $b_n \to 0$ because $\ln n \to \infty$. The Alternating Series Test gives convergence: the series converges CONDITIONALLY. It converges even more slowly than the alternating harmonic series.

$|a_n| = \left(1 + \frac{1}{n}\right)^n \to e$ (chapter 17). Since $e \neq 0$, $a_n$ does not tend to $0$: its even terms approach $e$ and its odd terms $-e$. The series diverges by the Divergence Test. The Alternating Series Test is irrelevant here, and checking whether $\left(1 + \frac{1}{n}\right)^n$ decreases (it actually increases) would be wasted time: look at the limit of $b_n$ FIRST, it is the cheapest hypothesis to check and the only one that can prove divergence.

$b_n = \frac{1}{n}$: positive, decreasing since $n + 1 > n$, and $\frac{1}{n} \to 0$. By the Alternating Series Test the series converges. The series of absolute values is $\sum \frac{1}{n}$, the harmonic series, a $p$-series with $p = 1$, which diverges. So the series converges conditionally, and this is the example to quote whenever a counterexample to converges implies converges absolutely is needed.