Sum of the tails of a convergent series If $s$ is a positive integer, consider the infinite sum 
\begin{align*}
\Sigma(s) = \sum_{m \geqslant 1} \sum_{n \geqslant m} \frac{1}{n^s},
\end{align*}
so that $\Sigma (s)$ is the sum of the tails of the series defining the Riemann zeta function $\zeta(s)$. Wolfram Alpha suggests that one has $\Sigma (s) = \zeta (s-1)$ for $s \geqslant 3$. Is this true, and (if yes) how can one prove this?
A more general question: Suppose $\sum_{n \geqslant 1} a_n$ converges to some complex number $A$. Under what conditions will the series
\begin{align*}
\sum_{m \geqslant 1} \sum_{n \geqslant m} a_n
\end{align*}
converge? If it is convergent, can anything be said about its value in terms of $A$? 
 A: Let $ s> 2 \cdot $
For $ \left(n,m\right)\in\left(\mathbb{N}^{*}\right)^{2} $, let's denote $ u_{n,m}\left(s\right)=\left\lbrace\begin{aligned}\frac{1}{n^{s}},\ \textrm{If }n\geq m\\ 0,\ \ \ \textrm{If }n<m\end{aligned}\right. $
Let's fixe $ n\in\mathbb{N}^{*} $, since $ \sum\limits_{m\geq 1}{u_{n,m}\left(s\right)} $ converges, $ \sum\limits_{m=1}^{+\infty}{u_{n,m}\left(s\right)}=\sum\limits_{m=1}^{n}{\frac{1}{n^{s}}}=\frac{1}{n^{s-1}} $, and $ \sum\limits_{n\geq 1}{\frac{1}{n^{s-1}}} $ converges.
$ \left(u_{n,m}\left(s\right)\right)_{\left(n,m\right)\in\left(\mathbb{N}^{*}\right)^{2}} $ is a summable family, and Fubini's theorem allows us to write that : $$ \sum_{m=1}^{+\infty}{\sum_{n=1}^{+\infty}{u_{n,m}\left(s\right)}}=\sum_{n=1}^{+\infty}{\sum_{m=1}^{+\infty}{u_{n,m}\left(s\right)}}=\sum_{n=1}^{+\infty}{\frac{1}{n^{s-1}}}=\zeta\left(s-1\right) $$
I suppose we could do pretty much the same thing to answer your second question :
If $ \sum\limits_{n\geq 1}{n a_{n}} $ do converges, we can, thanks to Fubini's theorem, write the following : $$ \sum_{m=1}^{+\infty}{\sum_{n=m}^{+\infty}{a_{n}}}=\sum_{n=1}^{+\infty}{n a_{n}} $$
A: The following is essentially the same as CHAMSI's answer, but I hope it's easier to read because it uses less notation.
Any term $1/n^s$ in the original series for $\zeta(s)$ occurs in exactly $n$ of the tails, namely those corresponding to $m=1,2,\dots,n$. When you add up all the tails, these $n$ copies of $1/n^s$ add up to $n/(n^s)=1/n^{s-1}$. So the sum of all the tails contains $1/n^{s-1}$ for each term $1/n^s$ in the original series for $\zeta(s)$. So the sum of the tails is $\zeta(s-1)$. 
The assumption that $s\geq3$ provides the (absolute) convergence needed to rearrange the order of summation implicit in the preceding paragraph.
