Suppose $\sum_{n=1}^\infty \frac{a_n}{n^2}< \infty$. Prove that $ \lim\limits_{n\to\infty} \frac{s_n}{n^2}= 0$. (a) Let $(a_n)^\infty_{n=1}$ be a sequence of non-negative real numbers such that
$\sum_{n=1}^\infty \frac{a_n}{n^2}< \infty$. Let $(s_n)^\infty_{n=1}$ be a sequence defined for all $n \in \mathbb{N} $ by $ s_n=\sum_{k=1}^n a_k$. Prove that
$ \lim\limits_{n\to\infty}  \frac{s_n}{n^2}= 0$.
(b)Use the sequence $a_n =\frac{n}{(1 + \log n)}$ to prove that the converse of part
(a) does not hold in general.
I tried in the  following way:
There is a theorem, If $\sum a_n$ converges then $\lim\limits_{n\to\infty} a_n =0$. I want to use the comparison test to show $\sum s_n/n^2$ converges.
$$\frac{s_n}{n^2}=\frac{\sum_{k=1}^n a_k}{n^2} \leq \sum_{n=1}^\infty \frac{a_k}{n^2} < \infty$$
So, $\sum s_n/n^2$ converges and hence, $ \lim\limits_{n\to\infty} \frac{s_n}{n^2}= 0$.
On the other hand, for $a_n=\frac{n}{1+\log n}$, $\sum \frac{a_n}{n^2}$ diverges if $\int_{1}^\infty \frac{a_n}{n^2}$ diverges.
\begin{align}
& \int_1^M \frac{a_n}{n^2}= \int_1^M \frac{1}{x(1+ \log x)} \, dx \\[6pt]
= {} & \log u \Big\vert_1^{1+\log M}=\log(1+ \log M )-\log 1 \\[6pt]
= {} & \log(1+\log M) \to \infty \text{ as } M \to \infty.
\end{align}
 A: For any $\varepsilon>0$, there is $N$ such that $0\leq \sum_{n> N}\frac{a_n}{n^2}\leq \varepsilon$
For all such $n$,
$$\frac{s_n}{n^2}=\frac{s_N}{n^2}+\frac{\sum^n_{m>N}a_m}{n^2}\leq \frac{s_N}{n^2}+\sum^n_{m>N}\frac{a_m}{m^2}<\frac{s_N}{n^2}+\varepsilon$$
Letting $n\rightarrow\infty$ gives
$\limsup_n\frac{s_n}{n^2}\leq\varepsilon$ for all $\varepsilon>0$. The conclusion follows from here.
A: Given $\epsilon>0$, there exists $N_1$ such that
$$ A = \sum_{n=N_1+1} ^ \infty \frac{a_n}{n^2} < \epsilon .$$
Then there exists $N_2$ such that
$$ B = \frac1{N_2^2} \sum_{n=1}^{N_1} a_n < \epsilon .$$
And if $N > \max\{N_1, N_2\}$, then
$$ \frac{s_N}{N^2} \le A + B .$$
A: Proof. (a) For every $\epsilon >0$, there exists $k$ such that $$\sum_{m>k}\frac{a_m}{m^2}<\epsilon.$$ It follows that for any $n>k,$ one has $$\sum_{m=k+1}^n \frac{a_m}{n^2}\leq \sum_{m=k+1}^n\frac {a_m}{m^2}<\epsilon$$ $$\Rightarrow \frac{s_n-s_k}{n^2}<\epsilon$$
$$\Rightarrow \limsup_{n\rightarrow \infty}\frac {s_n}{n^2}\leq \lim_{n\rightarrow \infty}\left(\frac {s_k}{n^2}+\epsilon\right)=\epsilon.$$ Since $\epsilon$ is arbitrary, one has $$\lim_{n\rightarrow\infty}\frac{s_n}{n^2}=0.$$
(b) To show that the converse is not true, you have checked that for $a_n=\frac n{1+\log(n)}$, $\sum_{n=1}^\infty\frac{a_n}{n^2}$ diverges. It remains to show that $$\lim_{n\rightarrow \infty}\frac {s_n}{n^2}=0.$$ Observe that $$\left(\frac x{1+\log(x)}\right)'=\frac{\log(x)}{(1+\log(x))^2}\geq 0,~{\rm for~}x\geq 1.$$ It follows that $$s_n=\sum_{m=1}^na_m\leq \int_1^{n+1}\frac x{1+\log(x)}~dx=\frac{N(n)}{1+\log(N(n))}\cdot n,\quad (1)$$ where $1\leq N(n)\leq n+1$ (by the mean value theorem for integral). One now shows that $N(n)\rightarrow \infty$ as $n\rightarrow \infty.$
Lemma 1. For $x\geq 1,$ one has $\sqrt{2x}\geq 1+\log(x).$
Proof. $$\sqrt{2x}\geq 1+\log(x)$$
$$\Leftrightarrow 2x\geq (1+\log(x))^2,$$ the latter is true if the function $f(x):=2x-(1+\log(x))^2$ satisfies $f(1)\geq 0$ and $f'(x)\geq 0$ for all $x\geq 1.$ Clearly $f(1)=1>0$ and $f'(x)=2-2(1+\log(x))\cdot \frac 1 x.$ Now $f'(x)\geq 0$ for $x\geq 1$ is equivalent to $x\geq 1+\log(x)$ for $x\geq 1,$ which is true since $g(x):=x-1-\log(x)$ satisfies $g(1)\geq 0$ and $g'(x)\geq 0$ for $x\geq 1.$
To prove the assertion that $N(n)\rightarrow \infty$ as $n\rightarrow \infty,$ one uses (1) and Lemma 1: $$\int_1^{n+1}\frac x{\sqrt{2x}}~dx\leq \int_1^{n+1}\frac x{1+\log(x)}~dx=\frac{N(n)}{1+\log(N(n))}\cdot n$$
$$\Rightarrow \frac{\sqrt{2}}3((n+1)^{3/2}-1)\leq \frac{N(n)}{1+\log(N(n))}\cdot n$$
$$\Rightarrow \frac{\sqrt{2}}3\cdot \frac{(n+1)+(n+1)^{1/2}+1}{(n+1)^{1/2}+1}\leq \frac{N(n)}{1+\log(N(n))}\leq N(n),$$ which shows that $N(n)\rightarrow \infty$ as $n\rightarrow \infty.$ Using this, one concludes from (1) that $$s_n\leq \frac{N(n)}{1+\log(N(n))}\cdot n\leq \frac{(n+1)n}{1+\log(N(n))}$$
$$\Rightarrow \frac {s_n}{n^2}\leq \frac 1{1+\log(N(n))}\cdot\frac{n+1}{n},$$ hence $$\lim_{n\rightarrow \infty}\frac {s_n}{n^2}=0,$$ as required.
