Implications of Convergence of a Series

Question

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?

Answer 1

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.

Answer 2

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). $B\leq \sum_{n/W<k\leq n}(a_k/k^2)\leq S(n/W)\leq S(M_2)\leq S(M_1)<\epsilon /2.$

Answer 3

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 $0.$