Exercises · 6. Series | Real Analysis

Easy

1Easy

Prove that $\sum_{n=2}^{\infty}\frac{\ln n}{n^2}$ converges, using the limit comparison test.

\frac{\ln n / n^2}{1/n^{3/2}} = \frac{\ln n}{\sqrt{n}} \to 0

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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 $p = \frac{3}{2} > 1$ .
$\frac{a_n}{b_n} = \frac{\ln n}{\sqrt{n}}$ . L'Hôpital's rule as $x \to \infty$ 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.

2Easy

For $N \ge 2$ , evaluate the partial product $P_N = \prod_{n=2}^{N}\left(1-\frac{1}{n^2}\right)$ and prove $\lim_{N\to\infty} P_N = \frac{1}{2}$ .

P_N = \frac{N+1}{2N} \to \frac{1}{2}

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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 $\frac{1}{2}$ .

3Easy

Show that $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)}$ converges, compute the third partial sum $S_3$ , and bound the remainder $|S - S_3|$ .

S_3 = \frac{1}{2}-\frac{1}{6}+\frac{1}{12} = \frac{5}{12}, \qquad |S-S_3| \le \frac{1}{20}

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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 $\frac{5}{12}$ .

Medium

4Medium

Explain why the ratio test is inconclusive for $\sum_{n=1}^{\infty} \frac{(2n)!}{4^n (n!)^2}\cdot\frac{1}{n}$ , then prove convergence with Raabe's test.

n\left(1-\frac{a_{n+1}}{a_n}\right) = \frac{n(3n+2)}{2(n+1)^2} \to \frac{3}{2} > 1

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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 $n\left(1-\frac{a_{n+1}}{a_n}\right) = \frac{n(3n+2)}{2(n+1)^2} \to \frac{3}{2}$ .
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 $a_n \sim \frac{1}{\sqrt{\pi}\,n^{3/2}}$ .)

5Medium

Prove that $\sum_{n=2}^{\infty}\frac{(-1)^n}{\,n+(-1)^n\sqrt{n}\,}$ converges conditionally. (Hint: use $\frac{1}{1+u} = 1-u+\frac{u^2}{1+u}$ .)

\frac{(-1)^n}{n+(-1)^n\sqrt{n}} = \frac{(-1)^n}{n}-\frac{1}{n^{3/2}}+\sigma_n, \quad |\sigma_n| \le \frac{C}{n^2}

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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 hint's identity 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.

Hard

6Hard

For real $p$ , prove by applying the condensation test twice that the Bertrand series $\sum_{n=3}^{\infty}\frac{1}{n\,\ln n\,(\ln\ln n)^p}$ converges if and only if $p > 1$ .

\sum 2^n a_{2^n} = \sum_{n\ge 3}\frac{1}{\ln 2\cdot n\,(\ln(n\ln 2))^p} \asymp \sum_{n\ge 3}\frac{1}{n(\ln n)^p}

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Put $a_n = \frac{1}{n\,\ln n\,(\ln\ln n)^p}$ ( $n \ge 3$ ). For $n \ge 3$ , $\ln\ln n > 0$ and each of $n$ , $\ln n$ , $(\ln\ln n)^p$ is positive and increasing, so $a_n$ decreases monotonically to $0$ . The condensation test applies.
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). 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 $0 < p \le 1$ .
For $p \le 0$ : if $n \ge 16$ then $\ln\ln n \ge 1$ , 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 $p > 1$ .

7Hard

Using the Euler product, prove that the infinite product $\prod_{p\ \text{prime}}\left(1+\frac{1}{p^2}\right)$ converges and equals $\frac{15}{\pi^2}$ . (You may use $\zeta(2) = \frac{\pi^2}{6}$ and $\zeta(4) = \frac{\pi^4}{90}$ as known.)

\prod_{p\le x}\left(1+\frac{1}{p^2}\right) = \frac{\prod_{p\le x}(1-p^{-2})^{-1}}{\prod_{p\le x}(1-p^{-4})^{-1}} \to \frac{\zeta(2)}{\zeta(4)} = \frac{15}{\pi^2}

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