I am trying to evaluate this sum, I know that $\sum\limits_{n=1}^\infty \dfrac{1}{n^2+n}$ is called telescopic series:
$$\sum_{n=1}^\infty \frac{1}{n(n+1)}$$
and I can show that as:
$$\frac{1}{k}-\frac{1}{k+1}$$ I would like to get some hint how I can evaluate it. Thanks!
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
Well, $$\sum_{n=1}^N\frac{1}{n(n+1)}=1-\frac{1}{N+1}$$ by telescopy, since, as you state $$\frac{1}{n(n+1)}=\frac 1n-\frac 1{n+1}$$
ADD A telescopic series is one of the form $$\sum x_n$$ where $x_n=y_{n+1}-y_n$ for some sequence. It follows that $$\sum_{n=1}^N x_n=\sum_{n=1}^Ny_{n+1}-\sum_{n=1}^N y_n\\=y_{N+1}+\underbrace{\sum_{n=1}^{N-1}y_{n+1}-\sum_{n=2}^N y_n}_{=0 \text{ Why? }}-y_1\\y_{N+1}-y_1$$
Thus $$\sum x_n=\lim_{N\to\infty} y_{N+1}-y_1$$ and the series converges if and only if $\lim\limits_{n\to\infty} y_n$ exists.
-
$\begingroup$ So if I evaluate his limit its 1? so most of the telescopic series are equal to 1? I'm just learning this so correct me if I'm wrong $\endgroup$ Commented Jul 1, 2013 at 18:52
-
-
$\begingroup$ What you noted a little hard to understand for me at least, I'll try to go over it a few more times but thanks anyway. what I`m understand it, its I can evaluate his limit ( of telescopic series? ). $\endgroup$ Commented Jul 1, 2013 at 19:00
-
$\begingroup$ Peter, re: last comment: oops, false alarm. :-( I see you've capped for the "day." We'll be sure you get there within a couple hours' span yet! ;-) $\endgroup$– amWhyCommented Jul 1, 2013 at 21:13
$\begingroup$
$\endgroup$
Hint: Try writing a small partial sum completely, and see what you're left with. Try to generalize the partial sum up to $N$, and then take the limit.
