How do I split this into partial sums? Given the series, $\sum(s_n - s_{n+1})$, 
can someone please explain to me how to split this into partial sums and the basic concept behind it?
I know that I can split it into $\sum s_n - \sum s_{n+1}$ but I don't know how to split it into partial sums.
 A: You may write, for the partial sum:
$$
\begin{align}
\sum_{n=0}^N(s_n-s_{n+1})&=\sum_{n=0}^Ns_n-\sum_{n=0}^Ns_{n+1}\\\\
&=\sum_{n=0}^Ns_n-\sum_{n=1}^{N+1}s_{n}\\\\
&=s_0+\sum_{n=1}^Ns_n-\sum_{n=1}^{N}s_{n}-s_{N+1}\\\\
&=s_0-s_{N+1}
\end{align}
$$ that what is called 'telescoping'.
A: This is a telescoping series.
Just notice that the $n$-th partial sum is
$$S_n=\sum_{k=1}^n (s_k-s_{k+1}) = (s_1-s_2)+ (s_2-s_3) + \dots + (s_{n-1}-s_n)+(s_n-s_{n+1})= s_1-s_{n+1},$$
since all remaining terms cancel out.
(If you want a more formal proof, you can show this by induction.)

EDIT: The OP asked in a comment and also in another question (which are now both deleted) how absolute convergence of $\sum (s_k-s_{k+1})$ relates to convergence of $(s_n)$ and $\sum s_n$.

If $\sum_{k=1}^\infty (s_k-s_{k+1})$ is absolutely convergent then it is also convergent.
Since the partial sums are equal to $s_1-s_{n+1}$ (see another question posted by the same user), this implies that the limit $\lim\limits_{n\to\infty} (s_1-s_{n+1})$ exists. Consequently, the sequence $s_n$ is convergent.

it seems possible that the OP might be interested not in convergence of the sequence $(s_n)$ but of the series $\sum s_n$.
Convergence of $\sum_{k=1}^\infty |s_k-s_{k+1}|$ does not implies convergence of $\sum_{k=1}^\infty s_k$.
To see this it suffices to notice that if we add arbitrary constant $c$ to the terms of sequence $(s_n)$, we get the same sum, since
$$|(s_k+c)-(s_{k+1}+c)|=|s_k-s_{k+1}|.$$
Therefore it is possible to choose $c$ in such way that $(s_k+c)$ is not convergent and therefore the series $\sum(s_k+c)$ does not converge either.
