Computing the sum of an infinite series I am confused as to how to evaluate the infinite series $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n^2+n}}.$$
I tried splitting the fraction into two parts, i.e. $\frac{\sqrt{n+1}}{\sqrt{n^2+n}}$ and $\frac{\sqrt{n}}{\sqrt{n^+n}}$, but we know the two individual infinite series diverge. Now how do I proceed?
Hint:

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

This kind of series is called Telescoping Series
Your problem may be converted to the following formula:
\begin{align}
&amp; \lim_{N\to\infty}\sum_{n=1}^{N}\left({\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}\right) =
\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{N+1}}\right) \\
&amp; \hspace{5mm} =
\lim_{N\to\infty}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{N+1}}\right)=
\lim_{N\to\infty}\left(\frac{\sqrt{N+1}-\sqrt{2}}{\sqrt{2}\sqrt{N+1}}\right) = 1
\end{align}