Convergence of the series $\sum k^2|a_k|$ If
$$\sum_{k>n}|a_k|=O\left( \frac{1}{n^\beta}\right)$$
for some $\beta>2$. I need to get that
$$\sum_{k=1}^{\infty}k^2|a_k|<\infty$$
I have a bit of a problem working with big $O$ notation, and I'm not getting an idea to solve this problem, because the information I have is about the "tail" of the series, not the series.
Edit: Using Greg Martin's idea I believe I solved the problem, but the resolution was long.
But I thought of another solution that I will post.
 A: It is possible to solve this problem by rearranging the sum,
\begin{matrix}
|a_1| & + & |a_2| & + & |a_3| & + & |a_4| & + & |a_5| & + & |a_6| & + & \cdots \\ 
 &  &  3|a_2| & + & 3|a_3| & + & 3|a_4| & + & 3|a_5| & + & 3|a_6| & +  & \cdots \\ 
 &  &   & & 5|a_3| & + & 5|a_4| & + & 5|a_5| & + & 5|a_6| & +  & \cdots \\ 
 &  &   & & & & 7|a_4| & + & 7|a_5| & + & 7|a_6| & +  & \cdots \\ 
 &  &   & & & &  &  & 9|a_5| & + & 9|a_6| & +  & \cdots \\ 
 &  &   & & & &  &  &  &  & \ddots  & \vdots & \cdots
\end{matrix}
so it's possible to rewrite the series,
$$\sum_{k \ge 1} k^2 | a_k | = \sum_{j \ge 1} [j^2 - (j-1)^2] \sum_{k \ge j} | a_k | = \sum_{j \ge 1} [2j-1] \sum_{k \ge j} | a_k |$$
as $\sum_{k>n}|a_k|=O\left( 1/n^\beta \right)$, exist $N$ and $c$ such that,
$$\sum_{k>n}|a_k|< \frac{c}{n^\beta}\qquad\text{ for all }n\ge N$$
thus,
$$\sum_{j \ge 1} [2j-1] \sum_{k \ge j} | a_k | = \sum_{j=1}^{N-1} [2j-1] \sum_{k \ge j} | a_k | + \sum_{j \ge N} [2j-1] \sum_{k \ge j} | a_k |$$
the first summation is a finite sum, so it is only necessary to check if the second summation has a finite sum.
$$\sum_{j \ge N} [2j-1] \sum_{k \ge j} | a_k | \le \sum_{j \ge N} [2j-1] \frac{c}{j^\beta} \le 2c  \sum_{j \ge N} \frac{j}{j^\beta}$$
as $\beta>2$ the series converges.
