Rudin: Problem Chp3.11 and need advice. I am working on the following problems and I have a couple of questions.

Suppose $a_n>0, s_n = \sum_{i = 1}^{n}$ and $\Sigma a_n$ diverges.
RTP
(a)  $\Sigma \frac{a_n}{1+a_n}$ diverges.
(b) By showing $$\sum_{i=1}^{k} \left[ \frac{a_{N+i}}{s_{N+i}} \right] \ge 1 - \frac{s_N}{s_{N+k}}$$
prove that $\sum \frac{a_n}{s_n}$ diverges.
(c) By showing  $$\frac{a_n}{s_n^2} \le \frac{1}{s_{n-1}}-\frac{1}{s_n}$$
and deduce that $\sum \frac{a_n}{s_n^2}$ converges.
(d) What can be said about 
$$\Sigma \frac{a_n}{n a_n +1} \text{ and } \Sigma \frac{a_n}{a+n^2a_n}$$

I understand (a).
But I am generally very iffy with proving these kind of problems because I usually get stuck with showing some inequalities being true.
1), How are the rest of the problem solved ?
2), Is there a general way to solve most of these inequalities ? Especially with those that have summations ?
3), Are these problem considered easy ? I am teaching myself analysis and it's such a struggle for me. 
 A: I was stuck on (b) until I realized that the inequality is misstated: it should be
$$\sum_{i=1}^k\frac{a_{N+i}}{s_{N+i}}\ge 1-\frac{s_N}{s_{N+k}}\;.$$
To see this, note that $\sum_{i=1}^ka_{N+i}=s_{N+k}-s_N$, and the partial sums are increasing, so
$$\sum_{i=1}^k\frac{a_{N+i}}{s_{N+i}}\ge\sum_{i=1}^k\frac{a_{N+i}}{s_{N+k}}=\frac1{s_{N+k}}\sum_{i=1}^ka_{N+i}=\frac{s_{N+k}-s_N}{s_{N+k}}=1-\frac{s_N}{s_{N+k}}\;.$$
By hypothesis $\lim\limits_{n\to\infty}s_n=\infty$, so for any fixed $N\in\Bbb Z^+$ we have $$\lim_{k\to\infty}\left(1-\frac{s_N}{s_{N+k}}\right)=1\;.$$
Thus, for each $N\in\Bbb Z^+$ there is a $k(N)\in\Bbb Z^+$ such that $k(N)>N$ and
$$\sum_{n=N+1}^{k(N)}\frac{a_n}{s_n}\ge\frac12\;.$$
Let $N_1=1$, and for $i\in\Bbb Z^+$ let $N_{i+1}=k(N_i)$. Then for each $m\in\Bbb Z^+$ we have
$$\sum_{n=1}^{N_m}\frac{a_n}{s_n}=\frac{a_1}{s_1}+\sum_{i=1}^{m-1}\sum_{n=N_i+1}^{N_{i+1}}\frac{a_n}{s_n}\ge 1+(m-1)\cdot\frac12=\frac{m+1}2\to\infty\quad\text{as}\quad m\to\infty\;,$$
and $\sum_{n\ge 1}\frac{a_n}{s_n}$ does indeed diverge.
I’d say that this one (in the corrected version) is of medium difficulty. Proving the inequality is the harder part, since its purpose should be fairly apparent: $\sum_{i=1}^k\frac{a_{N+i}}{s_{N+i}}$ is clearly the sum of the segment of the series from the $(N+1)$-st term through the $(N+k)$-th term, and the inequality says that we can bound it away from zero, so that we can prove divergence using the same technique that’s often used to show that the harmonic series is divergent.
For (c) the first thing that comes to mind is to combine the terms on the righthand side of the inequality into a single fraction to see what we get, and once that’s done, the rest turns out to be easy:
$$\frac1{s_{n-1}}-\frac1{s_n}=\frac{s_n-s_{n-1}}{s_{n-1}s_n}=\frac{a_n}{s_{n-1}s_n}\ge\frac{a_n}{s_n^2}\;,$$
since $s_{n-1}\le s_n$ and therefore $s_{n-1}s_n\le s_n^2$.
Summing terms like $\frac1{s_{n-1}}-\frac1{s_n}$ is easy, because they telescope, so let’s just replace the original series by an upper bound for it:
$$\begin{align*}
\sum_{n\ge 1}\frac{a_n}{s_n^2}&=\frac1{a_1}+\sum_{n\ge 2}\frac{a_n}{s_n^2}\\\\
&\le\frac1{a_1}+\sum_{n\ge 2}\left(\frac1{s_{n-1}}-\frac1{s_n}\right)\\\\
&=\frac1{a_1}+\frac1{s_1}-\frac1{s_2}+\frac1{s_2}-\frac1{s_3}+-\ldots\\\\
&=\frac1{a_1}+\frac1{s_1}\\\\
&=\frac1{a_1}\;.
\end{align*}$$
I did have to make a slight adjustment at the beginning to account for the lack of a non-zero $s_0$, but otherwise it was just using what was there in front of me; I’d classify this one as fairly easy.
For (d) note that if $a_n=\frac1n$, the first series diverges, and the second converges. Since the harmonic series is kind of a borderline divergent series, this suggests that the first series might always diverge and the second always converge. However, my attempts to prove that the first series must diverge foundered on the unpredictable behavior of the sequence $\langle a_n:n\in\Bbb Z^+\rangle$, and I eventually realized that $\sum_{n\ge 1}a_n$ might diverge because because of a very sparse (but infinite) set of large terms, while the other terms were converging to $0$ very rapidly, something like this:
$$a_n=\begin{cases}
n,&\text{if }n=k!\text{ for some }k\in\Bbb Z^+\\
2^{-n},&\text{otherwise}\;.
\end{cases}$$
The first case of that definition ensures that $\sum_{n\ge 1}a_n$ diverges. Now we have
$$\frac{a_n}{na_n+1}=\frac1{n+\frac1{a_n}}=\begin{cases}
\frac1{n+\frac1n},&\text{if }n=k!\text{ for some }k\in\Bbb Z^+\\\\
\frac1{n+2^n},&\text{otherwise}\;.
\end{cases}$$
$\sum_{n\ge 1}\frac1{n+2^n}$ is certainly convergent, and
$$\sum_{n\ge 1}\frac1{n!+\frac1{n!}}\le\sum_{n\ge 0}\frac1{n!}=e\;,$$
so in this case $\sum_{n\ge 1}\frac{a_n}{na_n+1}$ converges. Thus, we cannot in general say anything about the first series.
The second series, on the other hand, succumbs to the comparison test:
$$\frac{a_n}{a+n^2a_n}=\frac1{\frac{a}{a_n}+n^2}<\frac1{n^2}\;,$$
so it converges. (I am assuming here that $a>0$; it’s actually $1$ in my edition.)
Open-ended questions are always a bit harder, but I’d say that the second part of (d) isn’t bad. The first part, on the other hand, is probably the hardest part of the whole question.
A: A similar question where you could find the solution also applicable to your question: If sequence $a_n>0$ is decreasing to $0$ and $\sum_{k=1}^\infty a_n=\infty$, prove $\sum_{n=1}^\infty\frac{a_n}{e^{S_n}}<\infty$
The only difference is that it replace $S_n^2$ in the denominator by $e^{S_n}$. Actually it could be $f(S_n)$ where $\frac{1}{f(x)}$ is integrable over $[0,\infty)$. In your case $f(x)=x^2$ and in their case $f(x)=e^x$.
