Solution to Understanding Analysis 2nd Edition Exercise of Chapter 2.7

Exercise 2.7.2

(a) Converges. Since $\sum\limits_{n=0}^\infty \dfrac{1}{2^n}$ converges, see Example 2.7.5 (Geometric Series), and $\dfrac{1}{2^n+n}\leqslant \dfrac{1}{2^n}$, by Theorem 2.7.4 (Comparison Test), we have $\sum\limits_{n=0}^\infty \dfrac{1}{2^n+n}$ converges.

(b) Converges. Note that $\sum\limits_{n=0}^\infty \dfrac{1}{n^2}$ converges and $|\sin(n)|\leqslant 1$, by Theorem 2.7.4 (Comparison Test), we have $\sum\limits_{n=0}^\infty \dfrac{|\sin (n)|}{n^2}$ converges.  Then by Theorem 2.7.6 (Absolute Convergence Test) $\sum\limits_{n=0}^\infty \dfrac{\sin (n)}{n^2}$ converges as well.

(c) Diverges. The series is $\sum\limits_{n=1}^\infty \dfrac{n}{2n-2}$. Since $\lim \dfrac{n}{2n-2}=\dfrac{1}{2}\ne 0$, by Theorem 2.7.3, the series diverges.

Exercise 2.7.3

Exercise 2.7.4

(a) Let $x_n=\dfrac{1}{n}$, then $\sum x_n$ is the harmonic series and diverges. Let $y_n=(-1)^n$, then by Theorem 2.7.3 that $\sum y_n$ diverges. However $\sum x_ny_n$ is the alternating harmonic series which is convergent.

(b) Let $x_n=\dfrac{(-1)^n}{n}$, then $\sum x_n$ is the alternating harmonic series and converges. Let $y_n=(-1)^n$, then $\sum x_ny_n$ is the harmonic series which is convergent.

(c) Impossible. Note that $y_n=(x_n+y_n)-x_n$, by Theorem 2.7.1, if $\sum (x_n+y_n)$ and $\sum$ converge, then $\sum y_n$ converges as well.

(d) Let $x_{2n}=\dfrac{1}{n}$ and $x_{2n-1}=0$ for all $n\in\mathbf N$. Then $0\leqslant x_n\leqslant \dfrac{1}{n}$, and\[\sum x_n=\sum \frac{1}{n}\] is the harmonic series which is divergent.

Exercise 2.7.5

Exercise 2.7.6

(a) False. Let $a_n=1$ for all $n\in\mathbf N$, then the partial sum $s_n=n$. It is clear that no subsequence of $(s_n)$ converges. Hence $\sum a_n$ does not subverge.

(b) True. If $\sum a_n$ converges, so does the sequence of partial sums $(s_n)$. Hence by Theorem 2.5.2. any subsequence of $(s_n)$ converges to the same limit. Hence $\sum a_n$ subverges.

(c) True. Note that the partial sum $s_n$ of the sequence $\sum |a_n|$ is increasing. Since $\sum |a_n|$ subverges, there exists a subsequence of $(s_n)$ is convergent. Because $(s_n)$ is increasing, by Exercise 2.5.2 (d), $(s_n)$ converges. Hence $\sum |a_n|$ converges too. By Theorem 2.7.6 (Absolute Convergence Test), $\sum a_n$ converges.

(d) False. Let $a_{2n}=n$ and $a_{2n-1}=-n$, then the partial sum $s_{2n}$ satisfying \[s_{2n}=0.\]Therefore $\sum a_n$ subverges. However, no subsequences of $(a_n)$ are bounded. Hence no subsequences of $(a_n)$ are convergent.

Exercise 2.7.7

(a) Since $\lim (na_n)=l$, then there exists $N\in\mathbf N$ such that for all $n>N$ we have\[|na_n-l|<\frac{l}{2}\Longrightarrow na_n>\frac{l}{2}.\]Let $M=\min\{a_1,2a_2,\cdots,Na_N,l/2\}$. Since $a_n>0$ and $l>0$, we have $M>0$. Moreover, we have $na_n\geqslant M$ for all $n\in\mathbf N$. Hence \begin{equation}\label{}a_n\geqslant \frac{M}{n}.\end{equation}Since $\sum \dfrac{1}{n}$ diverges, we have $\sum \dfrac{M}{n}$ diverges (why?). Hence by Theorem 2.7.4 (Comparison Test) and \eqref{}, $\sum a_n$ diverges.

(b) Since $\lim (n^2a_n)$ exists, the sequence $(n^2a_n)$ is bounded by Theorem 2.3.2. Namely, there exists a $M>0$ such that \begin{equation}\label{}0 < n^2a_n\leqslant M\Longrightarrow 0< a_n<\frac{M}{n^2}.\end{equation}Since $\sum \dfrac{1}{n^2}$ converges, we have $\sum \dfrac{M}{n^2}$ converges (why?). Hence by Theorem 2.7.4 (Comparison Test) and \eqref{}, $\sum a_n$ converges.

Remark: the condition $a_n>0$ is not important because by Cauchy Criterion we only need to care about what happens when $n$ is large. Here the assumption $a_n>0$ simplifies the presentation of proof.

Exercise 2.7.8

(a) Since $\sum a_n$ converges absolutely, $\lim|a_n|=0$ by Theorem 2.7.3. Hence $(|a_n|)$ is bounded, so there exists $M>0$ such that $|a_n|\leqslant M$ for all $n\in\mathbf N$.

Since $\sum M|a_n|$ converges (Theorem 2.7.1 Algebraic Limit Theorem for Series ) and $M|a_n|\geqslant a_n^2$, by Theorem 2.7.4 (Comparison Test), $\sum a_n^2$ converges.

(b) False. Consider $a_n=b_n=\dfrac{(-1)^n}{\sqrt{n}}$. Since $\dfrac{1}{\sqrt{n}}$ is decreasing and $\lim \dfrac{1}{\sqrt{n}}=0$, by Theorem 2.7.7 (Alternating Series Test), $\sum a_n$ converges. Clearly $\lim b_n=0$. However, $\sum a_nb_n$ is the harmonic series which is divergent.

(c) True. We prove it by contradiction. Suppose $\sum n^2a_n$ converges, $\lim n^2a_n=0$ and hence $(n^2a_n)$ is bounded. Therefore there exists $M>0$ such that $|n^2a_n|\leqslant M$, namely\[|a_n|\leqslant \frac{M}{n^2}.\]Since  $\sum \dfrac{M}{n^2}$ converges, by Theorem 2.7.4 (Comparison Test), $\sum |a_n|$ converges and hence $\sum a_n$ converges absolutely which contradicts the assumption that $\sum a_n$ converges conditionally. Hence we are done.

Exercise 2.7.9

(a) Since $\lim \left|\dfrac{a_{n+1}}{a_n}\right|  = r <1 $, hence there exits $N\in\mathbf N$ such that \[\left| \left|\dfrac{a_{n+1}}{a_n}\right| -r\right|<r’-r\] for all $n\geqslant N$. Hence we have for all $n \geqslant N$,\[ \left|\dfrac{a_{n+1}}{a_n}\right| -r <r’-r\Longrightarrow \left|\dfrac{a_{n+1}}{a_n}\right|<r’.\]Then for all $n\geqslant N$, we have  \begin{equation}\label{}\left|\dfrac{a_{n+1}}{a_n}\right| < r’ \Longrightarrow | a_{n+1} |\leqslant r’ |a_n| \end{equation}for all $n> N$.

(b) Since $|r’|< 1$,, we have $\sum (r’)^n$ converges (geometric series). By Algebraic Limit Theorem for Series, $|a_N|\sum (r’)^n$ converges too.

(c) We show $\sum |a_n|$ is Cauchy. Given $\varepsilon > 0$, since $\sum (r’)^n$ converges, there exists $N_1$ such that for all $n>m\geqslant N_1$, we have\begin{equation}\label{} (r’)^{m+1}+\cdots+(r’)^n < \frac{(r’)^N\varepsilon}{|a_N|}.\end{equation}Note that if $k\geqslant M\geqslant N$, by \eqref{}, we have \begin{equation}\label{}|a_{k}|\leqslant r’|a_{k-1}|\leqslant \cdots \leqslant (r’)^{k-N}|a_N|.\end{equation} Let $M=\max\{N,N_1\}$, then for all $n> m > N$, by \eqref{}, we have\begin{align*}|a_{m+1}|+\cdots+|a_n| \leqslant &~ (r’)^{m+1-N}|a_N|+\cdots+(r’)^{n-N}|a_N|\\ =&~\frac{|a_N|}{(r’)^N}\big((r’)^{m+1}+\cdots+(r’)^n\big) \\ \text{by }  \eqref{} \quad < &~ \frac{|a_N|}{(r’)^N}\frac{(r’)^N\varepsilon}{|a_N|}=\varepsilon.\end{align*}Hence $\sum |a_n|$ is Cauchy and converges by Theorem 2.7.2 (Cauchy Criterion for Series). Therefore $\sum a_n$ converges by Theorem 2.7.6 (Absolute Convergence Test).

Exercise 2.7.10

Recall Exercise 2.4.10 (b), it would be convenient to express the infinite product as \[\prod_{n=1}^\infty (1+a_n)=(1+a_1)(1+a_2)\cdots.\]

(a) Yes. It is clear that $a_n=\dfrac{1}{2^{n-1}}$. Since $\sum \dfrac{1}{2^{n-1}}$ converges, by Exercise 2.4.10 (b), $\prod  (1+a_n)$ is also convergent.

(b) Because the sequence of partial products $(p_m)$ is decreasing and positive, hence by Monotone Convergence Theorem, $(p_m)$ converges.

It converges to zero. It suffices to show the product of their reciprocals diverges. Express the reciprocal as follows,\[\frac{2}{1}\cdot \frac{4}{3}\cdot \frac{6}{5}\cdots=\prod_{n=1}^\infty (1+a_n).\]Then $a_n=\dfrac{1}{2n-1}$. Note that $\dfrac{1}{2n-1} > \dfrac{1}{2}\cdot \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, by Theorem 2.7.4 (Comparison Test), $\sum \dfrac{1}{2n-1} $ diverges as well. Hence by Exercise 2.4.10 (b),\[\frac{2}{1}\cdot \frac{4}{3}\cdot \frac{6}{5}\cdots\] diverges (namely tends to infinity). Therefore the original infinite product series converges to zero.

(c) Yes. In this case, we have $a_n=\dfrac{1}{4n^2-1} \leqslant \dfrac{1}{n^2}$. Since $\sum \dfrac{1}{n^2}$ converges, by Theorem 2.7.4 (Comparison Test), $\sum \dfrac{1}{4n^2-1}$ converges too. Hence by Exercise 2.4.10 (b), the infinite product series converges.

Exercise 2.7.11

Exercise 2.7.12

Exercise 2.7.13


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