Convergence of the series $\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ Could you help me to understand for which $x$ this series converge  $\displaystyle\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$?
 A: Note that $$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{\sqrt{n+1}-\sqrt{n}}{n^x}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}.$$
If $x>1/2$, then 
$$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\leq\frac{1}{n^x(2\sqrt{n})}=\frac{1}{2n^{x+\frac{1}{2}}}.$$
Since $\displaystyle\sum\frac{1}{2n^{x+\frac{1}{2}}}$ converges when $x>1/2$ (see p-series), by comparison test, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ converges. 
On the other hand, if $x\le 1/2$,
$$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\geq\frac{1}{n^x(\sqrt{n})}=\frac{1}{n^{x+\frac{1}{2}}}\geq\frac{1}{n}.$$
Since the harmonic series diverges, by comparison test again, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ diverges. 
A: We know that $\displaystyle\sum_{n=1}^\infty n^{-s}$ converges for $s>1$.
As pointed out by Norbert, $\sqrt{n+1}-\sqrt{n}= \frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}+\sqrt{n}}$.
Hence, $\displaystyle\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n^x}<\displaystyle\sum_{n=1}^\infty \frac{1}{2\times{n^{\frac{1}{2}+x}}}$.
hence, for convergence, we must have the exponent of n $>1$. Hence, we must have $x>\frac{1}{2}$.
