Exercise 2.7.1
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.1
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
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.2
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
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.7
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{2.7.7.1}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{2.7.7.1}, $\sum a_n$ 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{2.7.7.2}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{2.7.7.2}, $\sum a_n$ 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{2.7.9.1}\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{2.7.9.2} (r’)^{m+1}+\cdots+(r’)^n < \frac{(r’)^N\varepsilon}{|a_N|}.\end{equation}Note that if $k\geqslant M\geqslant N$, by \eqref{2.7.9.1}, we have \begin{equation}\label{2.7.9.3}|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{2.7.9.3}, 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{2.7.9.2} \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
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.11
Exercise 2.7.12
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.12
Exercise 2.7.13
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.14
Exercise 2.7.14
See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.13
