series in series Let $\sum_{n=1}^\infty a_n<\infty$ be a complex convergent series
is the series $\sum_{n=1}^\infty (a_n \cdot \sum_{m=n+1}^\infty a_m)$ calculable ?
thanks for any suggestion
 A: This is not convergent in general.


*

*If your series is absolutely convergent, then indeed your new series will be absolutely convergent as well, by comparison (this is straightforward). 

*Otherwise, it may not be. Consider for instance the alternating series given by the general term $a_n \stackrel{\rm def}{=} \frac{(-1)^n}{\sqrt{n}}$ for $n\geq 1$. We have that $\sum_{n=1}^\infty a_n$ converges, but
$$
\sum_{k=n+1}^\infty a_k = \frac{(-1)^{n+1}}{2\sqrt{n}} + o\left(\frac{1}{\sqrt{n}}\right) 
$$
(if I am not mistaken); from which you get
$$
\sum_{n=1}^\infty \sum_{k=n+1}^\infty a_n a_k =
-\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} \xrightarrow[n\to\infty]{} -\infty
$$
by comparison.
A: Assuming that both the following series are convergent
$$ \sum_{n\geq 1}a_n,\qquad \sum_{n\geq 1}a_n^2 $$
we have the following symmetry trick:
$$ \sum_{n\geq 1}a_n\sum_{m>n}a_m = \sum_{1\leq n < m}a_n a_m = \sum_{1\leq m < n}a_n a_m = \frac{1}{2}\left[\left(\sum_{n\geq 1}a_n\right)^2-\sum_{n\geq 1}a_n^2\right]. $$
