# 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 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 at 17:32

Your problem may be converted to the following formula: $$\lim_{N\to\infty}\sum_{n=1}^{N}\left({\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}\right)= \left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{N+1}}\right)= \lim_{N\to\infty}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{N+1}}\right)= \lim_{N\to\infty}\left(\frac{\sqrt{N+1}-\sqrt{2}}{\sqrt{2}\sqrt{N+1}}\right) = 1$$

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