Exercise 4.1
Solution:
(1) Converges conditionally. It is convergent by Alternating Series Test. But $\sum_{k=1}^\infty\dfrac{1}{\sqrt{k}}$ is divergent by Proposition 4.16.
(2) Converges absolutely. By Geometric Series Test and the fact that $\ln 4 > 1$.
(3) Diverges. Note that \[\sqrt{n+1}-\sqrt n=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} > \frac{1}{2\sqrt{n+1}}.\]Then use Proposition 4.16 and Comparison Test.
(4) Diverges. By $k^{\rm th}$-term test and the fact that $\lim \dfrac{k}{k+7}=1 \ne 0$.
(5) Converges absolutely. Use the Ratio test in Exercise 4.9.
(6) Diverges. By $k^{\rm th}$-term test.
Exercise 4.2
Solution: Let $s_n$ be the partial sum of the series $\sum_{k=1}^\infty a_k$. Let $S_n$ be the partial sum of the series $\sum_{k=1}^\infty |a_k|$.
Since the series $\sum_{k=1}^\infty |a_k|$ converges, the sequence $(S_n)$ converges and hence is Cauchy. For any $\e >0$, there exists $N$ such that $$|S_{m}-S_n|=\sum_{k=n+1}^m|a_k|$$for all $m>n > N$.By the triangle inequality, we have\[\left|\sum_{k=n+1}^ma_k\right|\leqslant \sum_{k=n+1}^m|a_k|.\]Therefore for all $m > n > N$, we have\begin{align*}|s_m-s_n|&=\left|\sum_{k=n+1}^ma_k\right|\\ &\leqslant \sum_{k=n+1}^m|a_k| < \e.\end{align*}Hence the sequence $(s_n)$ is also Cauchy and thus converges by Theorem 3.42. This shows that the series $\sum_{k=1}^\infty a_k$ converges.
Exercise 4.3
Solution: If $|r| > 1$, then let $|r|=1+a$ where $a >0$. We have\[|r|^n=(1+a)^n=1+ n a+\frac{n(n-1)}{2}a+\cdots > 1+na.\]Hence the sequence $(|r|^n)$ is not bounded, so is $(r^n)$. Hence $(r^n)$ diverges.
If $r=1$, it is clear that $(r^n)$ converges to one. If $r=-1$, it is clear that $(r^n)$ diverges.
If $|r| < 1$, we show that $(r^n)$ converges to $0$. Since $\dfrac{1}{|r|} > 1$, we set $\dfrac{1}{|r|}=1+a$, where $a > 0$.
For any $\e >0$, let $N=\dfrac{1}{a\e}$, then $$\frac{1}{na} < \e$$for all $n> N$. Recall from above that $(1+a)^ n > 1+na$, we have\[|r|^n=\frac{1}{(1+a)^n} < \frac{1}{1+na} < \frac{1}{na} < \e.\]Therefore $(r^n)$ converges to zero.
Exercise 4.4
Solution: Consider the harmonic series given by $a_n=\dfrac{1}{n}$. It is known from Proposition 4.15 that $\sum_{k=1}^\infty a_k$ diverges. But the series $$\sum_{k=1}^\infty a_k^2=\sum_{k=1}^\infty \frac{1}{k^2}$$ is convergent by the Series $p$-Test.
Exercise 4.5
Solution:
(a) Since $\sum_{k=1}^\infty a_k$ converges, the sequence $(a_n)$ converges to zero by $k^{\rm th}$-term Test. Hence $(a_n)$ is bounded by Proposition 3.20. Hence there exists $\alpha>0$ such that $a_k < \alpha$ for all $k$. In particular, we have $$0 < a_k^2 < \alpha a_k$$ for all $k$.
Since $\sum_{k=1}^\infty a_k$ converges, so does $\sum_{k=1}^\infty \alpha a_k$ converges. Since $a_k^2 \leqslant \alpha a_k$, by Proposition 4.12 (Comparison test), $\sum_{k=1}^\infty a_k^2$ converges.
(b) Consider the series $\sum_{k=1}^\infty (-1)^k\frac{1}{\sqrt k}$. This series is convergent by Exercise 1 (1), but $\sum_{k=1}^\infty a_k^2$ is the harmonic series which is divergent.
Exercise 4.6
Solution: Taking the series $\sum_{k=1}^\infty (-1)^k\frac{1}{\sqrt k}$ from Exercise 4.5 (b). Then $\sum_{k=1}^\infty a_k$ converges, but $\sum_{k=1}^\infty a_k^2$ diverges. It is easy to see using Alternating series Test that $\sum_{k=1}^\infty a_k^3$ converges.
Exercise 4.11
Solution: Let $\sum_{k=1}^\infty a_k$ be the harmonic series, then it is divergent. Moreover,\[\lim_{k\to \infty}(a_{k+1}-a_k)=\lim_{k\to \infty}\left(\frac{1}{k+1}-\frac{1}{k}\right)=0.\]
Exercise 4.12
Solution:
(1) Let $a_k=\dfrac{1}{k^2}$ and $b_k=\dfrac{1}{k}$. Then $0\leqslant a_k \leqslant b_k$, $\sum_{k=1}^\infty a_k$ converges, but $\sum_{k=1}^\infty b_k$ diverges.
(2) Let $a_k=\dfrac{1}{k^2}$ and $b_k=\dfrac{1}{k}$. Then $0\leqslant a_k \leqslant b_k$, $\sum_{k=1}^\infty b_k$ diverges, but $\sum_{k=1}^\infty a_k$ converges.
Exercise 4.13
Solution: Let $a_k=-\dfrac{1}{k}$. Since the harmonic series $\sum_{k=1}^\infty \dfrac{1}{k}$ diverges, the series $\sum_{k=1}^\infty a_k$ diverges.
Let $b_k=\dfrac{1}{k^2}$, by the $p$-Series Test, $\sum_{k=1}^\infty b_k$ converges. It is clear that $a_k \leqslant b_k$ for all $k$.
Exercise 4.14
Solution:
(a) This is exactly Exercise 3.19.
(b) Let $a_n=(-1)^{n-1}$. Then $\sum_{k=1}^n a_n$ diverges by $k^{\rm th}$-term Test. We have\[s_{2n-1}=1,\quad s_{2n}=0.\]Hence\[\frac{s_1+\cdots+s_n}{n}=\frac{[n/2]}{n}\]\[\Longrightarrow \frac{n/2-1}{n}\leqslant \frac{s_1+\cdots+s_n}{n} \leqslant \frac{1}{2}.\]Since $\displaystyle\lim_{k\to \infty}\dfrac{k/2-1}{k}=\dfrac{1}{2}$, using the Sequence Squeeze Theorem (Theorem 3.23), we conclude that $$\lim_{n\to \infty}\frac{s_1+\cdots+s_n}{n}=\frac{1}{2}.$$Hence $(a_n)$ is Cesaro summable but divergent.
