Value of sum of telescoping series $$\sum_{n\geqslant1}\frac{1}{\sqrt{n}} -\frac{ 1}{\sqrt {n+2}}$$
In looking at the first five partial sums, I am not convinced the series is telescopic (the middle terms don't cancel out).
Thanks in advance!
 A: $$\sum_{n=1}^N\left(\frac1{\sqrt n}-\frac1{\sqrt{n+2}}\right)=1-\frac1{\sqrt3}+\frac1{\sqrt2}-\frac1{\sqrt4}+\frac1{\sqrt3}-\frac1{\sqrt5}+\ldots=$$
$$=1+\frac1{\sqrt2}-\frac1{\sqrt{N+1}}-\frac1{\sqrt{N+2}}\xrightarrow[N\to\infty]{}\ldots ?$$
A: This is a telescopic series $\sum\limits_{n\geqslant1}\left(\frac1{\sqrt{n}}-\frac1{\sqrt{n+2}}\right)=\sum\limits_{n\geqslant1}(a_n-a_{n+1})$, for $a_n=\frac1{\sqrt{n}}+\frac1{\sqrt {n+1}}$. It converges because $a_n\to0$, and its sum is $a_1=1+\frac1{\sqrt2}$.

More generally, for  every integer $k\geqslant1$ and function $\varphi$, the series 
$$
\sum\limits_{n\geqslant1}\left(\varphi(n)-\varphi(n+k)\right)
$$ 
is telescopic, it converges when $\varphi(n)\to0$ when $n\to\infty$, then its sum is $\sum\limits_{n=1}^k\varphi(n)$.
A: $\sum_{k=1} ^n \frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k+2}}$
$= \frac{1}{\sqrt{1}}- \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{2}}- \frac{1}{\sqrt{4}} +\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{4}}- \frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{5}}- \frac{1}{\sqrt{7}}+...+ \frac{1}{\sqrt{n-2}}- \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n+1}}+ \frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+2}}$
the first five terms dont cancel immediatley, after many terms have been computed they cancel,i will not classify it under telescope because the terms dont entirely cancel
