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I have to check the convergence of this series:

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

Which is equal to $$\sum_{n=1}^\infty \frac{\sqrt{n+1}}{\sqrt{n}} - 1$$

What can I do here to check whether this series convergences or not? Thank you very much.

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    $\displaystyle{\,\sqrt{\, n + 1\,}\, - \,\sqrt{\,n\,}\, \over \,\sqrt{\,n\,}} \sim {1 \over 2n}\ \mbox{as}\ n \to \infty$. – 2017-01-23

2 Answers 2

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Hint. One has, for $n\ge1$, $$ {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}={\frac{1}{\sqrt{n} (\sqrt{n+1}+ \sqrt{n})}}\ge {\frac{1}{\sqrt{n+1} (\sqrt{n+1}+ \sqrt{n+1})}}=\frac{1}{2(n+1)} $$ then the initial series diverges as does the harmonic series.

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    I can't see why this diverges, the harmonic series $$ \frac{1}{n} $$ is > than $$ \frac{1}{2(n+1)} $$ – 2017-01-22
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    @Blnpwr Observe that $\sum_{n=1}^N\frac1{2(n+1)}=\frac12\sum_{k=2}^{N+1}\frac1{k}$ with the change of index $k=n+1$. So $\sum_{n=1}^N\frac1{2(n+1)}$ is also a partial sum of the harmonic series, it is just written differently. – 2017-01-22
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    ahh, I see, thank you very much. – 2017-01-22
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    @Blnpwr You are welcome! – 2017-01-22
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HINT:

Multiply numerator and denominator by $\sqrt{n+1}+\sqrt{n}$ to see that

$$\begin{align} \frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}&=\frac{1}{\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)}\\\\ &\ge \frac{1}{2(n+1)} \end{align}$$

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    Just curious ... since the posted answers are effectively the same, what was the basis for awarding the other with the best vote? – 2017-01-22