# 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!

• split it up into the sum of a series with odd $n$'s and a series with even $n$'s – Stefan Smith Nov 16 '13 at 1:59

## 3 Answers

$$\sum_{n=1}^N\left(\frac1{\sqrt n}-\frac1{\sqrt{n+2}}\right)=1-\frac1{\sqrt3}+\frac1{\sqrt2}-\frac1{\sqrt4}+\frac1{\sqrt3}-\frac1{\sqrt5}+\ldots=$$

$$=1+\frac1{\sqrt2}-\frac1{\sqrt{N+1}}-\frac1{\sqrt{N+2}}\xrightarrow[N\to\infty]{}\ldots ?$$

• Yes, though it also goes to zero. Editing on its way. – DonAntonio Nov 15 '13 at 16:29

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 $\sum\limits_{n=1}^k\varphi(n)$.

$\sum_{k=1} ^n \frac{1}{\sqrt{k}}- \frac{1}{\sqrt{k+2}}$

$= \frac{1}{\sqrt{1}}- \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{2}}- \frac{1}{\sqrt{4}} +\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{4}}- \frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{5}}- \frac{1}{\sqrt{7}}+...+ \frac{1}{\sqrt{n-2}}- \frac{1}{\sqrt{n}}+ \frac{1}{\sqrt{n-1}}- \frac{1}{\sqrt{n+1}}+ \frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+2}}$

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

• The second and fourth term cancel. Why do you think they don't? Only the first and the third one don't cancel. – DonAntonio Nov 15 '13 at 16:14
• so i shud jus say that $=1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{n+2}}$,but wolfram classifies it as harmonic $H^{\frac{1}{2}}$ – Jonas Kgomo Nov 16 '13 at 20:15