# 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?

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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

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And thank you!! – Eddy Aug 4 '14 at 17:33

Note that $\frac{\sqrt{n+1}}{\sqrt{n(n+1)}}=\frac1{\sqrt n}$ and $\frac{\sqrt{n}}{\sqrt{n(n+1)}}=\frac1{\sqrt {n+1}}$. KEyword: Telescope

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Thank you very much!! – Eddy Aug 4 '14 at 17:32

Your problem may be converted to the following formula: \begin{align} & \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) \\ & \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}

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I know this is an old question, but since Leucippus and MartinSleziak edited it, what happened to the first term in your series? You are missing the 1. – user46944 Jul 30 '15 at 16:26