2
$\begingroup$

$$\sum_{n\geqslant1}\frac{1}{\sqrt{n}} -\frac{ 1}{\sqrt {n+2}}$$

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!

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

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

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

$\endgroup$
  • $\begingroup$ Yes, though it also goes to zero. Editing on its way. $\endgroup$ – DonAntonio Nov 15 '13 at 16:29
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.