Exercises

Put $a_n = \frac{\ln n}{n^2} > 0$ ($n \ge 2$) and take $b_n = \frac{1}{n^{3/2}}$ as the comparison series. $\sum b_n$ converges since it is a $p$-series with $p = \frac{3}{2} > 1$.

$\frac{a_n}{b_n} = \frac{\ln n}{\sqrt{n}}$.

Verify L'Hôpital's hypotheses: as $x \to \infty$, $\ln x \to \infty$ and $x^{1/2} \to \infty$, so this is an $\frac{\infty}{\infty}$ form.

L'Hôpital's rule then gives $\frac{\ln x}{x^{1/2}} \to \frac{1/x}{\frac{1}{2}x^{-1/2}} = \frac{2}{\sqrt{x}} \to 0$, so $\frac{a_n}{b_n} \to 0$.

By the limit-$0$ case of the limit comparison test: if $\frac{a_n}{b_n} \to 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. Hence $\sum_{n=2}^{\infty}\frac{\ln n}{n^2}$ converges.

Since $1-\frac{1}{n^2} = \frac{n^2-1}{n^2} = \frac{(n-1)(n+1)}{n^2}$, split the product in two: $P_N = \prod_{n=2}^{N}\frac{n-1}{n}\cdot\prod_{n=2}^{N}\frac{n+1}{n}$.

The first product telescopes: $\frac{1}{2}\cdot\frac{2}{3}\cdots\frac{N-1}{N} = \frac{1}{N}$. So does the second: $\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{N+1}{N} = \frac{N+1}{2}$.

Hence $P_N = \frac{1}{N}\cdot\frac{N+1}{2} = \frac{N+1}{2N} = \frac{1}{2}+\frac{1}{2N} \to \frac{1}{2}$. So the infinite product $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^2}\right)$ converges to $\frac{1}{2}$.

With $b_n = \frac{1}{n(n+1)}$, $b_n$ decreases monotonically to $0$ because $n(n+1)$ is strictly increasing. Hence the series converges by the alternating series (Leibniz) test.

$S_3 = \frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} = \frac{1}{2}-\frac{1}{6}+\frac{1}{12} = \frac{6-2+1}{12} = \frac{5}{12}$.

The alternating-series error bound gives $|S - S_3| \le b_4 = \frac{1}{4\cdot 5} = \frac{1}{20} = 0.05$. So $S$ lies within $0.05$ of $\frac{5}{12}$.

Put $a_n = \frac{(2n)!}{4^n (n!)^2}\cdot\frac{1}{n} > 0$. Then $\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{4}\cdot\frac{1}{(n+1)^2}\cdot\frac{n}{n+1} = \frac{n(2n+1)}{2(n+1)^2} \to 1$, so the ratio test is inconclusive.

Compute the Raabe quantity: $1-\frac{a_{n+1}}{a_n} = \frac{2(n+1)^2-n(2n+1)}{2(n+1)^2} = \frac{3n+2}{2(n+1)^2}$, hence $n\left(1-\frac{a_{n+1}}{a_n}\right) = \frac{n(3n+2)}{2(n+1)^2} \to \frac{3}{2}$.

The chapter's Raabe form is $n\left(\frac{a_n}{a_{n+1}} - 1\right)$; the two limits agree: since $\frac{a_{n+1}}{a_n} \to 1$, $n\left(\frac{a_n}{a_{n+1}} - 1\right) = n\left(1 - \frac{a_{n+1}}{a_n}\right)\cdot\frac{a_n}{a_{n+1}}$ and the extra factor $\frac{a_n}{a_{n+1}} \to 1$, so both limits are the same.

Since the limit $\frac{3}{2} > 1$, Raabe's test gives convergence of $\sum a_n$.

(Indeed $\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$, so $a_n \sim \frac{1}{\sqrt{\pi}\,n^{3/2}}$.)

Put $b_n = \frac{1}{n+(-1)^n\sqrt{n}} > 0$ ($n \ge 2$).

Factoring $n$ from the denominator, $b_n = \frac{1}{n}\cdot\frac{1}{1+u_n}$ with $u_n = \frac{(-1)^n}{\sqrt{n}}$, $|u_n| \le \frac{1}{\sqrt{2}}$.

The bound gives $u_n \ge -\frac{1}{\sqrt{2}} > -1$, so $1 + u_n \ge 1 - \frac{1}{\sqrt{2}} > 0$; hence $u_n \ne -1$ and the hint's identity $\frac{1}{1+u_n} = 1-u_n+\frac{u_n^2}{1+u_n}$ is valid.

The identity therefore gives $\frac{1}{1+u_n} = 1-u_n+\rho_n$ where $|\rho_n| = \frac{u_n^2}{|1+u_n|} \le \frac{C}{n}$ with $C = \frac{1}{1-1/\sqrt{2}}$.

Hence $(-1)^n b_n = \frac{(-1)^n}{n}-\frac{(-1)^n u_n}{n}+\frac{(-1)^n\rho_n}{n}$, and since $(-1)^n u_n = \frac{1}{\sqrt{n}}$, we get $(-1)^n b_n = \frac{(-1)^n}{n}-\frac{1}{n^{3/2}}+\sigma_n$ with $|\sigma_n| \le \frac{C}{n^2}$. The three series $\sum\frac{(-1)^n}{n}$ (alternating series test), $\sum\frac{1}{n^{3/2}}$ ($p$-series), and $\sum\sigma_n$ (absolutely convergent) all converge, so $\sum (-1)^n b_n$ converges.

It does not converge absolutely: $n+(-1)^n\sqrt{n} \le n+\sqrt{n} \le 2n$ ($n \ge 1$), so $b_n \ge \frac{1}{2n}$, and $\sum\frac{1}{2n}$ diverges; by comparison $\sum b_n$ diverges too. Hence the convergence is conditional.

Put $a_n = \frac{1}{n\,\ln n\,(\ln\ln n)^p}$ ($n \ge 3$).

Treat only $p \ge 0$ by condensation first: for $n \ge 3$, $\ln\ln n > 0$, $n$ and $\ln n$ are positive and increasing, and for $p \ge 0$ so is $(\ln\ln n)^p$; hence $a_n$ decreases monotonically to $0$.

The condensation test applies.

(For $p < 0$, $(\ln\ln n)^p$ is decreasing and the monotonicity claim is false — e.g. for $p = -1$, $a_3 \approx 0.0285 < a_4 \approx 0.0589$. The case $p < 0$ is handled in the last step by comparison alone, with no condensation.)

First condensation: $2^n a_{2^n} = \frac{1}{\ln 2\cdot n\,(\ln(n\ln 2))^p}$. Since $\ln(n\ln 2) = \ln n+\ln\ln 2$ with $\ln\ln 2 \approx -0.3665$, we have $\frac{1}{2}\ln n \le \ln(n\ln 2) \le \ln n$ for $n \ge 3$ (the upper bound from $\ln 2 < 1$; the lower from $\frac{1}{2}\ln n \ge -\ln\ln 2$, true for $n \ge 3$).

For $p > 0$ these two-sided bounds make the condensed series comparable, up to constants, to $\sum_{n\ge 3}\frac{1}{n(\ln n)^p}$ (comparison test).

Re-verify the condensation hypotheses for the second application: for $p > 0$, $b_n = \frac{1}{n(\ln n)^p}$ is positive, and since $n$ and $(\ln n)^p$ are both positive and increasing, $b_n$ decreases monotonically to $0$ — so condensation applies again.

Condensing again gives $\sum 2^n\cdot\frac{1}{2^n(n\ln 2)^p} = \frac{1}{(\ln 2)^p}\sum\frac{1}{n^p}$, which converges only for $p > 1$.

Hence for $p > 0$ we get convergence when $p > 1$ and divergence when $0 < p \le 1$.

The case $p < 0$ (and $p = 0$) is handled by comparison alone — no monotonicity of the original series is needed: if $n \ge 16$ then $\ln\ln n \ge 1$ (since $n \ge e^e \approx 15.15$), so $(\ln\ln n)^p \le 1$ and $a_n \ge \frac{1}{n\ln n}$.

Since $\sum\frac{1}{n\ln n}$ diverges (one condensation turns it into a constant multiple of the harmonic series), comparison gives divergence of the original series too.

Thus convergence holds if and only if $p > 1$.

For $s > 1$ the Euler product $\zeta(s) = \prod_{p}(1-p^{-s})^{-1}$ holds (understood as the limit of finite partial products). The key identity is $1+\frac{1}{p^2} = \frac{1-p^{-4}}{1-p^{-2}}$ — division is legitimate since the denominator is nonzero ($p \ge 2$).

Over the finite set of primes $\{p \le x\}$, $\prod_{p\le x}\left(1+\frac{1}{p^2}\right) = \frac{\prod_{p\le x}(1-p^{-4})}{\prod_{p\le x}(1-p^{-2})} = \frac{\prod_{p\le x}(1-p^{-2})^{-1}}{\prod_{p\le x}(1-p^{-4})^{-1}}$. Every factor is positive, so no rearrangement issue arises.

As $x \to \infty$, the numerator tends to $\zeta(2) = \frac{\pi^2}{6}$ and the denominator to $\zeta(4) = \frac{\pi^4}{90}$ (Euler product).

The denominator's limit $\zeta(4) \ne 0$ — no factor of the convergent Euler product is zero — so the quotient limit law applies.

Hence the infinite product converges to $\frac{\pi^2/6}{\pi^4/90} = \frac{90}{6\pi^2} = \frac{15}{\pi^2}$.