Implications of Convergence of a Series Let us say that:
$$
\sum_{k = 1}^{\infty} \frac{a_k}{k^2} <\infty
$$
Where the $a_k \geq 0$. Can we say anything about what the limit:
$$
\lim_{n\to \infty} \sum_{k = 1}^{n}\frac{a_k}{n^2}
$$
converges to? I know the limit converges by the comparison test, but is there anything we can say about its value?
 A: Let $S_n = \sum_{k=1}^n \frac{a_k}{k^2}$ where $\lim_{n \to \infty}S_n = S$.
Summing by parts we get
$$\sum_{k=1}^n a_k =  \sum_{k=1}^n k^2 \frac{a_k}{k^2} = n^2S_n + \sum_{k=1}^{n-1}S_k \left(k^2 - (k+1)^2 \right),$$
and
$$\frac{1}{n^2}\sum_{k=1}^n a_k  = S_n - \frac{1}{n^2}\sum_{k=1}^{n-1}(2k+1)S_k$$
The first term on the RHS converges to $S$ and the second term on the RHS converges to $S$ by Stolz-Cesaro:
$$\lim_{n \to \infty}\frac{1}{n^2}\sum_{k=1}^{n-1}(2k+1)S_k = \lim_{n \to \infty}\frac{(2n+1)S_n}{(n+1)^2 - n^2} = \lim_{n \to \infty}S_n = S$$
Hence,
$$\lim_{n \to \infty} \frac{1}{n^2}\sum_{k=1}^n a_k = 0$$
and this is true whether or not the $a_k$ are always nonnegative.
A: Let $\epsilon>0.$ Then there exists $k_0$ such that $\sum_{k>k_0} \frac{a_k}{k^2} < \epsilon.$ For $n>k_0$ we then have
$$\sum_{k = 1}^{n}\frac{a_k}{n^2} = \sum_{k = 1}^{k_0}\frac{a_k}{n^2}\, + \sum_{k = k_0+1}^{n}\frac{a_k}{n^2} = \sum_{k = 1}^{k_0}\frac{a_k}{n^2} + \sum_{k = k_0+1}^{n}\frac{k^2}{n^2}\frac{a_k}{k^2} < \sum_{k = 1}^{k_0}\frac{a_k}{n^2} +\epsilon.$$
It follows that $\limsup_{n\to \infty} \sum_{k = 1}^{n}\dfrac{a_k}{n^2} \le \epsilon.$ Since $\epsilon$ was arbitrary, the desired limit is $0.$
A: The limit is $0$.
Let $S=\sum_{k=1}^{\infty}a_k/k^2.$ For brevity let $S(m)=\sum_{k=m}^{\infty} (a_k/k^2).$ 
Given $\epsilon >0,$ take $W>1$ such that $S/W^2<\epsilon /2.$ Take $M_1>1$ such that $ S(M_1))<\epsilon/2, $  and take  $M_2>WM_1.$
Let $\sum_{k=1}^n(a_k/n^2)=A+B$ where $A=\sum_{k\leq n/W}(a_k/n^2)$  and $B=\sum_{n/W<k\leq n}(a_k/n^2).$
Now for all $n>M_2$ we have:
(i). $k\leq n/W \implies (a_k/n^2)\leq (a_k/k^2)/W^2.$ Therefore $$A\leq (\sum_{k=1}^n a_k/k^2)/W^2\leq S/W^2<\epsilon /2.$$
(ii). $B\leq \sum_{n/W<k\leq n}(a_k/k^2)\leq S(n/W)\leq S(M_2)\leq S(M_1)<\epsilon /2.$ 
