Convergence of the series $\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$

Question

Could you help me to understand for which $x$ this series converge $\displaystyle\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$?

Answer 1

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.

Answer 2

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 $x>\frac{1}{2}$.