Problem 2.
(g) $\dis \sum_{n=1}^\infty \frac{\ln n}{n^p+1}$ converges iff (by limit comparison test) $\dis \sum_{n=1}^\infty \frac{\ln n}{n^p}$ iff (by integral test) $\dis \int_1^\infty \frac{\ln x}{x^p}\,dx$ converges. The last integral converges only when $p>1$, hence $\dis \sum_{n=1}^\infty \frac{\ln n}{n^p+1}$ converges.
The integral test applies because $\dis \frac{\ln x}{x^p} $ is decreasing on $[3,\infty)$. To show this, observe that for every $x\ge 3>e$, \[
\frac{d}{dx}\brac{ \frac{\ln x}{x^p}} =\frac{1-p\ln x}{x^{p+1}}<\frac{1-\ln x}{x^{p+1}} <\frac{1-\ln e}{x^{p+1}}=0.
\]
Remark. Unless the integrand is very simple to tell it is decreasing or not, otherwise try not to use integral test.
(j) Since \[
\lim_{x\to 0}\frac{\tan x}{x} =\lim_{x\to 0} \frac{\sin x}{x}\frac{1}{\cos x}=1\cdot 1=1,
\] hence when $x$ is small, $\tan x$ is very close to $x$. In other words, when $n$ is big, $\tan \frac{1}{n}$ is very close to $\frac{1}{n}$. Indeed, since \[
\limn \frac{\tan \frac{1}{n}}{\frac{1}{n}} = 1,
\] hence $\sum \tan \frac{1}{n}$ converges iff $\sum \frac{1}{n}$ converges by limit comparison test. As $\sum \frac{1}{n}$ diverges, so does $\sum \tan \frac{1}{n}$.
(k) and (l) depends on the following limit \[
\lim_{x\to 0}\frac{\ln (1+x)}{x}=1.
\] Which says that $\ln (1+x)$ is very close to $x$ when $x$ is small.
(k) Since $\dis \ln \frac{n+2}{n+1}=\ln \brac{1+\frac{1}{n+1}} \approx \frac{1}{n+1}$ when $n$ is large, indeed, \[
\limn \frac{\ln \brac{1+\frac{1}{n+1}}}{\frac{1}{n+1}}=1,
\] hence $\sum \frac{1}{n+1}$ diverges $\implies$ $\sum \ln \brac{1+\frac{1}{n+1}}$ diverges.
(l) This is essentially same as (k), the only difference is that each term $\ln \frac{n}{n+1}\leq 0$, however, $-\ln \frac{n}{n+1}=\ln \frac{n+1}{n}=\ln \brac{1+\frac{1}{n}}\ge 0$, hence by the method in (k), $\sum -\ln \frac{n}{n+1}$ diverges to $+\infty$, and thus $\sum \ln \frac{n}{n+1}$ diverges to $-\infty$.
Problem 3. Let $y=(\ln(\ln k))^{p\ln k}$, then \begin{align*}
\ln y&=p\ln k \ln\ln\ln k\\
y&=(e^{\ln k})^{p\ln\ln \ln k}\\
&= k^{p\ln\ln\ln k}.
\end{align*} Hence the original series becomes \[
\sum_{k=2}^\infty \frac{1}{k^{p\ln\ln\ln k}}.
\] It seems that any tests fail to test the convergence, however, this can be shown very easily by comparison test! Recall the slogan: 「分母越細個數越大」(the smaller the denominator, the larger the value). Let's try to shrink $k^{p\ln\ln\ln k}$ to $k^{2}$. Of course this shrinking is not true for any $k$, but it should be intuitive to see that when $k$ is large enough, $p\ln\ln\ln k \ge 2$, such that the shrinking is working. Indeed how large should $k$ be? Well, let's observe that \begin{align*}
\iff p\ln\ln\ln k &\ge 2\\
\iff\ln\ln\ln k &\ge \frac{2}{p}\\
\iff \ln\ln k &\ge \exp \frac{2}{p}\\
\iff \ln k &\ge \exp(\exp \frac{2}{p}) \\
\iff k &\ge \exp(\exp(\exp \frac{2}{p}))= e^{e^{e^{ 2/p}}}.
\end{align*} Hence if we let $N$ be an integer bigger than $\exp(\exp(\exp \frac{2}{p}))$, then \[
k\ge N\implies p\ln\ln\ln k \ge 2\implies \frac{1}{k^{p\ln\ln\ln k}}\leq \frac{1}{k^2},
\] hence $\dis \sum_{k=N}^\infty \frac{1}{k^{p\ln\ln\ln k}}$ converges by comparison test, and thus $\dis \sum_{k=1}^\infty \frac{1}{k^{p\ln\ln\ln k}}$ converges (finitely many terms do not affect the convergence of the whole series!).
Problem 4. (a) Consider the continuous functions with the following graph:
\sum_{n=1}^\infty \frac{1}{2n^2}=\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n^2}<\infty.
\] (b) Typo: In this problem the codomain of the function should be $\R$.
Let $s_n=1+1/2+1/3+\cdots +1/n$, then $\limn s_n=\infty$. Let $f:[0,\infty)\to \R$ be a function defined as follows: For $n=0,1,2,\dots$, \[
f(x)=\begin{cases}
1&\text{if }s_{2n}\leq x < s_{2n+1} \\
-1&\text{if } s_{2n+1}\leq x<s_{2n+2}
\end{cases}
\] The graph of this function is roughly as follows:
[(s_1-0) - (s_2-s_1)] + [(s_3-s_2) - (s_4-s_3)]+\cdots + [(s_{2k+1}-s_{2k})-(s_{2k+2}-s_{2k+1})]+\cdots
\] or more precisely, \[
\sum_{k=0}^\infty [(s_{2k+1}-s_{2k})-(s_{2k+2}-s_{2k+1})] = \sum_{k=0}^\infty\brac{\frac{1}{2k+1}-\frac{1}{2k+2}}.
\] It is left as an exercise to check $ \sum_{k=0}^\infty\brac{\frac{1}{2k+1}-\frac{1}{2k+2}}$ converges.
Problem 5. For the sake of contradiction, let's suppose $\{b_n\}$ is bounded, then there is an $M>0$ such that $\dis b_k=\frac{1}{a_{k+1}}-\frac{1}{a_k}\leq M$ for every $k\ge 1$. It follows that by taking $\sum_{k=1}^n$ on both sides, \[
\frac{1}{a_{n+1}}-\frac{1}{a_1}\leq nM,
\] and hence\begin{equation}\label{we get contra}
a_{n+1}\ge \frac{1}{nM+\frac{1}{a_1}}\implies\infty> \sum_{n=1}^\infty a_{n+1}\ge \sum_{n=1}^\infty \frac{1}{nM+\frac{1}{a_1}}.
\end{equation} However, when $n$ is very big, $\frac{1}{nM+\frac{1}{a_1}}\approx \frac{1}{nM}$, indeed, \[
\limn \frac{\dis \frac{1}{nM+\frac{1}{a_1}}}{\dis \frac{1}{nM}}=1,
\] hence $\sum \frac{1}{nM+\frac{1}{a_1}}$ and $\frac{1}{M}\sum \frac{1}{n}$ converges and diverges at the same time. As $\frac{1}{M}\sum \frac{1}{n}$ diverges, so does $\sum \frac{1}{nM+\frac{1}{a_1}}$, a contradiction to (\ref{we get contra}).
Problem 6. Let $p_n=(1+a_1)(1+a_2)\cdots (1+a_n)$, then for $n\ge 2$, $p_n/p_{n-1}-1=a_n$, hence \[
\frac{a_n}{(1+a_1)\cdots (1+a_n)}=\frac{p_n/p_{n-1}-1}{p_n}=\frac{1}{p_{n-1}}-\frac{1}{p_n}.
\] It follows that\begin{align*}
\sum_{n=2}^\infty \frac{a_n}{(1+a_1)\cdots (1+a_n)}& = \sum_{n=2}^\infty\brac{ \frac{1}{p_{n-1}}-\frac{1}{p_n}}\\
&=\lim_{k\to\infty} \sum_{n=2}^k \brac{ \frac{1}{p_{n-1}}-\frac{1}{p_n}}\\
&=\lim_{k\to\infty}\brac{ \frac{1}{p_1}-\frac{1}{p_k}}\\
&\leq \frac{1}{p_1}<\infty,
\end{align*}by divergence test, \[
\limn \frac{a_n}{(1+a_1)\cdots (1+a_n)}=0.
\]
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