# Evaluate series of $\sum_{n=1}^\infty \frac{1}{n^2+n}$

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!

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.

• 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 Commented Jul 1, 2013 at 18:52
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.