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Solution to Principles of Mathematical Analysis Chapter 3 Part A


Chapter 3 Numerical Sequences and Series

Exercise 1

(By ghostofgarborg) Assume $s_n \to s$. Choose $\epsilon > 0$, and let $N$ be such that $|s_n – s| < \epsilon$ when $n > N$. Then \[ ||s_n|-|s|| \leq |s – s_n| < \epsilon \]whenever $n > N$. This implies convergence of $|s_n|$.

The converse is not true. Let $s_n = (-1)^n$. This sequence does not converge, even though $|s_n|$ does.


Exercise 2

(By ghostofgarborg) Note that $ (\sqrt{n^2 + n} – n) = n/(\sqrt{n^2 + n} + n) $, and therefore
\[\frac {1}{2 + \frac{1}{\sqrt n}} =
\frac {n}{(n + \sqrt n) + n} <
\sqrt{n^2 + n} – n <
\frac{n}{n+n} =
\frac 1 2
\]where $(2 + 1/\sqrt n)^{-1} \to \frac 1 2$. By theorem 3.19, $ \lim_{n \to \infty} \sqrt{n^2+n} – n = \frac 1 2$.


(By Dan kyp44 Whitman) We can calculate this limit directly:
\[
\lim_{n \to \infty} (\sqrt{n^2 + n} – n)
= \lim_{n \to \infty} (\sqrt{n^2 + n} – n) \left(\frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} + n}\right) \\
\]\[
= \lim_{n \to \infty} \frac{n^2 + n – n^2}{\sqrt{n^2 + n} + n}
= \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + n} + n}
= \lim_{n \to \infty} \frac{1}{\sqrt{1 + 1/n} + 1}
= \frac{1}{\sqrt{1} + 1}
= \frac{1}{2} \,,
\]noting that clearly $\sqrt{n^2 + n} + n > 0$.


Exercise 3

(By ghostofgarborg) Note that if $s_n < 2$, $s_{n+1} = \sqrt{2 + \sqrt{s_n}} < \sqrt{4} = 2 $, so the sequence is bounded. Also note that if $s_{n-1} > s_{n-2}$, then $s_n = \sqrt{2 + \sqrt{s_{n-1}}} > \sqrt{2 + \sqrt{s_{n-2}}} = s_{n-1}$, so the sequence is monotonically increasing. It therefore converges by thm 3.14.


Exercise 4

(By ghostofgarborg) We observe that if $a_n = (2^n – 1)/2^n$ and $b_n = (2^{n-1} -1)/2^n$, then $s_{2n+1} = a_n$ and $s_{2n} = b_n$ fulfills both the initial condition and the recursion, so that this is a closed form of the sequence. Note that $a_n \to 1$, $b_n \to \frac 1 2$, and that any subsequence of $s_n$ contains either a subsequence of $a_n$ or a subsequence of $b_n$. Consequently, any convergent subsequence converges to either $1$ or $\frac 1 2$. We get \[ \limsup_{n \to \infty} s_n = 1 \qquad \liminf_{n \to \infty} s_n = \frac 1 2.\]


Exercise 5

(By ghostofgarborg) If $\limsup a_n = \infty$ and $\limsup b_n > -\infty$, the result is trivial. If $\limsup a_n = – \infty$ and $\limsup b_n < \infty$, then $b_n$ is bounded above by $B$ and given a number $N$ \[ \# \{ n : a_n + b_n > N \} \leq \# \{ n : a_n > N – B \} \]the latter of which is finite. Therefore $\limsup (a_n + b_n) = -\infty = \limsup a_n + \limsup b_n$.

We can therefore assume that both $\limsup a_n$ and $\limsup b_n$ are finite. There is a subsequence such that $a_{n_k} + b_{n_k} \to \limsup (a_n + b_n)$. By choosing this subsequence right, i.e. if necessary passing to another subsequence, we can make sure that $ a_{n_k} $ is convergent. Then $ b_{n_k} = (a_{n_k} + b_{n_k}) – a_{n_k}$ is convergent, so
\[ \limsup_n (a_n + b_n) = \lim_k (a_{n_k} + b_{n_k}) = \lim_k a_{n_k} + \lim_k b_{n_k} \leq \limsup_n a_n + \limsup_n b_n. \]


Exercise 6

(By ghostofgarborg)

(a) Since the series telescopes, $S_n = \sum_{i=0}^n (\sqrt{i+1} – \sqrt i) = \sqrt{n+1}$. This shows that it diverges to $\infty$.

(b) We have\[ a_n = \frac{1}{n(\sqrt {n+1} + \sqrt n)} < \frac{1}{n \sqrt n}. \]Converges by the comparison test (thm. 3.25).

(c) We have \[ \limsup_{n \to \infty} \sqrt[n]{|a_n|} = \limsup_{n \to \infty} (\sqrt[n]{n} – 1) = 1 – 1 = 0 \]Converges by the root test.

(d) When $|z| \leq 1$, \[ \left| \frac{1}{1 + z^n} \right| \geq \frac{1}{1+|z|^n} \geq \frac 1 2 \]violating the necessary condition that $a_n \to 0$. In this case, $\sum a_n$
is divergent. When $|z| > 1$, \[ \left| \frac{1}{1 + z^n} \right| \leq \frac{1}{|z|^n – 1} \leq \frac{1}{|z|^n – |z|^{n-1}} = \left( \frac{|z|}{|z|-1} \right) \frac{1}{|z|^n} \] and $\sum a_n$ converges by comparison with the geometric series $\frac {|z|}{|z|-1} \sum \frac{1}{|z|^n}$.


Exercise 7

(By ghostofgarborg) Assume $\sum a_n = C$. We know that $\sum \frac{1}{n^2}$ converges. Let $D$ be the number it converges to. Note that $ S_n = \sum_{i=0}^n \sqrt{a_i} / i$ is monotonic. By the Cauchy-Schwarz inequality \[ \sum_{i=1}^{n} \frac{\sqrt{a_i}}{i} \leq \left( \sum_{i=1}^{n} a_i \right)^{\frac 1 2} \left( \sum_{i=0}^{n} \frac{1}{i^2} \right)^{\frac 1 2} < \sqrt{C \cdot D} \]This implies that $\sum \frac{\sqrt{a_n}}{n}$ is bounded, hence converges.


Exercise 8

(By ghostofgarborg) We first note that thm 3.42 holds for $b_n$ a monotonously increasing sequence whose limit is 0 as well, since $(-b_n)$ then fulfills the criteria of the theorem, and $\sum a_nb_n = – \sum a_n(-b_n)$.

If $\sum a_n$ converges, the partial sums form a bounded sequence. If $b_n$ is monotonic and bounded it converges to a number $B$, and we get that \[ \sum a_n b_n = \sum a_n(b_n – B) + B \sum a_n \]The first sum on the right hand side converges by thm 3.42 and the observation above. The second sum converges because $\sum a_n$ does. Consequently the left hand side converges.


Exercise 9

(By ghostofgarborg)

(a) We have\[ \frac 1 R = \lim_{n \to \infty} \frac{(n+1)^3}{n^3} = \lim_{n \to \infty} ( 1 + \frac 3 n + \frac 3 {n^2} + \frac 1 {n^3} ) = 1 \]So $R = 1$.

(b) We have \[ \frac 1 R = \lim_{n \to \infty} \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \lim_{n \to \infty} \frac 2 {n+1} = 0 \]So $R = \infty$.

(c) We have \[ \frac 1 R = \lim_{n \to \infty} \sqrt[n]{\frac{2^n}{n^2}} = \lim_{n \to \infty} \frac{2}{(\sqrt[n]{n})^2} = 2 \] So $R = \frac 1 2$.

(d) We have \[ \frac 1 R = \lim_{n \to \infty} \sqrt[n]{\frac{n^3}{3^n}} = \lim_{n \to \infty} \frac{(\sqrt[n]{n})^3}{3} = \frac 1 3 \] So $R = 3$.


Exercise 10

(By ghostofgarborg) If $|z| \geq 1$ and $a_n \neq 0$ for an infinite number of values of $n$, then $|a_n z^n| \geq |a_n|$, and does not tend to $0$. This makes the series divergent when $|z| \geq 1$.


Exercise 11

(By ghostofgarborg)

(a) If $a_n \leq 1$, then $\frac{a_n}{1+a_n} \geq \frac{a_n}{2}$. If $a_n > 1$, then $\frac{a_n}{1+a_n} > \frac{1}{2}$. Therefore \[ \sum \frac{a_n}{1+a_n} \geq \frac 1 2 \sum \min(a_n, 1) \] If $ \{ n : a_n > 1 \} $ is infinite, this clearly diverges. Otherwise, there is an $N$ such that $ n > N $ implies $a_n \leq 1$, in which case the series diverges by comparison to $\frac 1 2 \sum a_n$.

(b) Since $S_n$ is monotonically increasing, whenever $ j \leq k $ \[ \frac{a_{n+j}}{S_{n+j}} \geq \frac{a_{n+j}}{S_{n+k}}.\]Consequently, \[ \frac{a_{N+1}}{S_{N+1}} + \cdots + \frac{a_{N+k}}{S_{N+k}} \geq \frac{a_{N+1}}{S_{N+k}} + \cdots + \frac{a+{N+k}}{S_{N+k}}
= \frac{a_{N+1} + \cdots + a_{N+k}}{S_{N+k}}
= 1 – \frac{S_{N}}{S_{N+k}}\]Since $S_n$ is increasing and unbounded, it is possible to choose $k$ such that $S_{N}/S_{N+k}$ is arbitrarily close to $0$. This shows that $\frac{a_n}{S_n}$ is not Cauchy, and therefore not convergent.

(c) Since $S_n$ is monotonically increasing, we have that $ S_n^2 \geq S_{n-1}S_n $, so we can deduce that \[ \frac{a_n}{S_n^2} \leq \frac{S_n – S_{n-1}}{S_{n-1}S_n} = \frac{1}{S_{n-1}} – \frac{1}{S_{n}} \] Since $1/S_n$ is bounded \[ \sum_{n=1}^N \frac{a_n}{S_n^2} \leq \frac{a_1}{S_1^2} + \sum_{n=2}^N \left( \frac 1 {S_{n-1}} – \frac 1 {S_n} \right) \leq \frac{a_1}{S_1^2} + \frac 1 {S_{1}} – \frac 1 {S_N} \] so $\sum a_n/S_n^2$ is monotonic and bounded, hence convergent.

(d) Since \[ \frac{a_n}{1 + n^2a_n} \leq \frac{1}{n^2} \] $\sum a_n/(1+n^2a_n)$ converges. The series $\sum a_n/(1+na_n)$, however, might converge or diverge. Let $a_n = \frac 1 n$, and it is clear that it diverges. Let $a_n = 1$ whenever $n$ is a square and $a_n = 2^{-n}$ otherwise. This series clearly diverges, since the terms do not tend to 0 as $n \to \infty$. Then \[ \sum_{n=1}^N \frac{a_n}{1+na_n} \leq \sum_{n=1}^N \frac{1}{n^2} + \sum_{n=1}^{N} \frac 1 {2^n} \] and the series therefore converges.


Exercise 12

(By ghostofgarborg)

(a) Since $r_n$ is monotonically decreasing \[ \frac{a_m}{r_m} + \cdots + \frac{a_n}{r_n} > \frac{a_{m}}{r_m} + \cdots + \frac{a_{n-1}}{r_m} = \frac{a_{m} + \cdots + a_{n-1}}{r_m} = \frac{r_{m} – r_n}{r_m} = 1 – \frac{r_n}{r_m} \] Since $r_n \to 0$, given any $M$ we can find an $N > M$ such that $1 – \frac {R_N}{R_M} $ is arbitrarily close to $1$. This implies that $a_n/r_n$ is not Cauchy, hence not convergent.

(b) We have \[ \frac{a_n}{\sqrt{r_n}} = \frac{a_n \left( 1 + \frac{\sqrt{r_{n+1}}}{\sqrt{r_n}} \right)}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} < \frac{2a_n}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} = \frac{2(r_{n} – r_{n+1})}{\sqrt{r_{n}} + \sqrt{r_{n+1}}} = 2(\sqrt{r_{n}} – \sqrt{r_{n+1}}) \]Since \[ \sum_{n=1}^N 2(\sqrt{r_{n}} – \sqrt{r_{n+1}}) = 2(\sqrt {r_{1}} – \sqrt{r_{N+1}}) < 2\sqrt{r_1}\]the series converges by the comparison test.


Exercise 13

(By ghostofgarborg) Since $ \sum |a_n|$ and $ \sum |b_n|$ converge, so does their Cauchy product $ \sum C_n$. Let $c_n$ be the Cauchy product of $a_n$ and $b_n$. Then $|c_n| \leq C_n$, and as it is majorized by a convergent series, $\sum |c_n|$ converges . I.e. $\sum c_n$ converges absolutely.


Exercise 14

(By ghostofgarborg)

(a) Choose $\epsilon > 0$ and $N$ such that $|s_n – s_m| < \epsilon/2$ for all $n, m > N$. Then choose $M$ such that \[ \left| \frac{(s_0 – s) + \cdots + (s_{N-1} – s)}{M} \right| < \frac \epsilon 2 \] Then whenever $n > \max(M, N)$ \[ \left| \frac{(s_0 – s) + \cdots + (s_n-s)}{n+1} \right| \leq \left| \frac{(s_0 – s) + \cdots + (s_{N-1} – s)}{n+1} \right| + \left| \frac{(S_N-s) + \cdots + (S_n-s)}{n+1} \right| < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon \]This shows that $(\sigma_n -s)$ converges to $0$, i.e. that $\sigma_n \to s$.

(b) Let $s_n = (-1)^n$. Then $s_n$ does not converge, but $\sigma_n = \frac{1 + (-1)^{n}}{2(n+1)}$ converges to $0$.

(c) Yes. Let $a_n = \frac 1 {2^n}$, $b_{2^n} = n$, and $ b_n = 0$ when $n$ is not a power of $2$.
Let $s_n = a_n + b_n$. Then $s_n > 0$, and the series is unbounded, so $\limsup s_n = \infty$. However\[ 0 < \sigma_n = \frac 1 {n+1} \sum_{i=0}^n \frac {1}{2^i} + \frac{1}{n+1} \sum_{i=0}^{\lfloor \log_2(n) \rfloor} i < \frac{2}{n} + \frac{\log_2(n)^2}{n} \]where the right hand side tends to 0. This shows that $\sigma_n \to 0$.

(d) Note that\[ \sum_{k=1}^n k(s_k – s_{k-1}) \ = \ \sum_{k=1}^n ks_k – \sum_{k=0}^{n-1} (k+1)s_k \ = \ ns_n – \sum_{k=0}^{n-1}s_k \ = \ (n+1)s_n – \sum_{k=0}^n s_k \]so that
\[ s_n – \sigma_n = \frac{1}{n+1} \sum_{k=0}^n k a_k \] Note that the right hand side is the $n$th arithmetic mean of the sequence $na_n$, which by assumption and our result in (a) converges. Therefore $s_n – \sigma_n$ is convergent, and since $\sigma_n$ is convergent by assumption, $(s_n – \sigma_n) + \sigma_n = s_n$ is convergent.

(e) We have \begin{align*}
s_n – \sigma_n &= s_n – \sum_{k=0}^n \frac{s_k}{n+1} \\
&= s_n – \sum_{k=0}^n \left(\frac{1}{n-m} – \frac{m+1}{(n+1)(n-m)} \right) s_k \\
&= \sum_{k=0}^m \left(\frac{(m+1)s_k}{(n+1)(n-m)} – \frac{s_k}{n-m} \right) + \sum_{k=m+1}^n \left(\frac{(m+1)s_k}{(n+1)(n-m)} + \frac{s_n – s_k}{n-m} \right) \\
&= \frac{m+1}{n-m} \left( \sum_{k=0}^n \frac{s_k}{n+1} – \sum_{k=0}^m \frac{s_k}{m+1} \right) + \frac{1}{n-m} \sum_{k=m+1}^n (s_n – s_k)
\end{align*} which is the desired equality. By assumption $|s_n – s_{n-1}| < M/n$, so that \[ |s_n – s_t| = | (s_n – s_{n-1}) + (s_{n-1} – s_{n-2}) + \cdots + (s_{i+1} – s_i)| \leq \frac M n + \cdots + \frac {M}{i+1} \leq \frac{(n-i)M}{i+1} \]The rest of the argument is covered in sufficient detail in the text.

Baby Rudin 数学分析原理完整第三章习题解答

Linearity

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