# comparison test of $\sum\frac{\sqrt{n+1}-\sqrt{n}}{n}$

I want to know if the sum $$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$ converges or not. I've tried the ratio and root test but they don't fit. So wolframalpha says that the sum converges by the comparison test. So I've tried to find a convergent majorizing sum (I've tried e.g. $\sum\frac1{n^\alpha}$ with $\alpha>1$ etc.) but I can't find one.

does anybody know one?

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## 1 Answer

$$\frac{\sqrt{n+1}-\sqrt n}{n}=\frac{1}{n(\sqrt{n+1}+\sqrt n)}<n^{-\frac32}$$

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Note that between the first and second steps there's an implicit 'clearing of the numerator' by multiplying by $\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$ and then using $\left(\sqrt{n+1}+\sqrt{n}\right)\cdot\left(\sqrt{n+1}-\sqrt{n}\right) = \left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2=n+1-n=1$. – Steven Stadnicki Dec 19 '12 at 17:22