Check Convergence of the series: $\sum_{n=1}^\infty {\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}}$ 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. 
 A: 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.
A: HINT:
Multiply numerator and denominator by $\sqrt{n+1}+\sqrt{n}$ to see that
$$\begin{align}
\frac{ \sqrt{n+1} - \sqrt{n}}{\sqrt{n}}&=\frac{1}{\sqrt{n}\left(\sqrt{n+1}+\sqrt{n}\right)}\\\\
&\ge \frac{1}{2(n+1)}
\end{align}$$
A: Write the partial sum as
$$s_m = \sum_{n=1}^{m}\left( \sqrt{1+\frac{1}{n}}-1\right)$$
For $n\ge 1$ we have the inequality
$$\left(\frac{1}{n}+1\right)-\left(-\frac{1}{8 n^2}+\frac{1}{2 n}+1\right)^2=\frac{8 n-1}{64 n^4}>0$$
whence by taking the square root
$$\sqrt{\frac{1}{n}+1}>-\frac{1}{8 n^2}+\frac{1}{2 n}+1\\
\sqrt{\frac{1}{n}+1}-1>-\frac{1}{8 n^2}+\frac{1}{2 n}
$$
Hence
$$s_m \gt \sum_{n=1}^{m}\left( -\frac{1}{8 n^2}+\frac{1}{2 n}\right)\\
\gt \frac{1}{2}H_m-\frac{\pi^2}{48}$$
Summary: $s_m$ diverges logarithmically like the harmonic number.
