how to write the general formula for a telescoping series Im trying to check if the series 
$\sum_{k=2}^{\infty} \frac{1}{\sqrt{k-1}} - \frac{1}{\sqrt{k+1}} $
is converging or not.
I have divided the series into two parts with the series with even k >= 2 and the series with the odd k >= 3.
For the series with even k >=2 I have started writing down the terms:
$S_k = (1-\frac{1}{\sqrt{3}})+ (\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}})+(\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}})+...$
But I am wodering how I can write down the last two terms (k-1, and k) to show that this series converges. 
 A: To remain on the safe side of strictness consider the partial sum
$$s(n) = \sum_{k=2}^n \left(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k+1}}\right)$$
and then look for the limit $n\to\infty$.
We have
$$s(n) = \left( \frac{1}{\sqrt{1}} +\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+ ...+ \frac{1}{\sqrt{n-1}}\right)\\
-\left( \frac{1}{\sqrt{3}} +\frac{1}{\sqrt{4}}+ ...+ \frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\\
= 1+\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}  $$
and we get finally
$$\lim_{n\to \infty } \, s(n) =1+\frac{1}{\sqrt{2}} $$
EDIT
The generalization to an arbitrary sequence $\{a(k)\}$ is easily done. For a fixed distance $d$ and for $n>d$ the partial sum is given by
$$s(d,n) = \sum_{k=1}^n \left( a_{n}-a_{k+d} \right)= \sum_{j=1}^d a_j-\sum_{j=1}^d a_{n+j}$$
And if $\lim_{n\to \infty } \, a_n =0$ we find
$$s(d) = \lim_{n\to \infty } \, s(d,n) =\sum_{j=1}^d a_j$$
A: Since$$\sum_{k=2}^\infty\left(\frac1{\sqrt{k-1}}-\frac1{\sqrt{k+1}}\right)=\sum_{k=2}^\infty\left(\frac1{\sqrt{k-1}}-\frac1{\sqrt k}\right)+\sum_{k=2}^\infty\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right),$$the sum of your series is $1+\dfrac1{\sqrt2}$.
