# Determine for which $x$ the sum $\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ converges

My attempt:

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

From here, I've tried using the ratio test, but haven't been able to simplify it in a clever way. I can bound the above sum by $\sum_{n=1}^\infty\frac{1}{n^x}$, but this won't show all $x\in\mathbb{R}$ that cause the sum to converge. Basically, I am stuck now.

Any help appreciated!

• For which $x$ does $\sum \frac{1}{n^x\sqrt{n}}$ converge? – Daniel Fischer Feb 6 '17 at 20:32

HINT:

$$\frac{1}{n^x\left(\sqrt{n+1}+\sqrt{n}\right)}< \frac{1}{n^x(2\sqrt{n})}=\frac12 \frac{1}{n^{x+1/2}}$$

and

$$\frac{1}{n^x\left(\sqrt{n+1}+\sqrt{n}\right)}> \frac{1}{n^x(3\sqrt{n})}=\frac13 \frac{1}{n^{x+1/2}}$$

• So for these examples, do we have that $\frac{1}{2}\frac{1}{n^{x+1/2}}$ converges for $x>1/2$ and $\frac{1}{3}\frac{1}{n^{x+1/2}}$ diverges for $x<1/2$, so the sum converges for all $x>1/2$? – Jess Feb 7 '17 at 12:58
• @jess Yes. The series converges when $x>1/2$ and diverges elsewhere. – Mark Viola Feb 7 '17 at 15:15

As this is a series with positive terms, I would use equivalents:

$\sqrt{n+1}-\sqrt{n}\sim_\infty\dfrac1{2\sqrt n}$, so $$\frac{\sqrt{n+1}-\sqrt{n}}{n^x}\sim_\infty\dfrac1{2\, n^{\,x+\frac12}},$$ which converges if and only if $x+\frac12>1$, i.e. $x>\frac12$, and diverges if and only if $x\le \frac12$.

• You probably meant $n^{x+1/2}$ instead of $\sqrt{n}^{x+1/2}$ – kingW3 Feb 6 '17 at 21:02
• Oh! yes. Thanks for pointing it. Fixed now. – Bernard Feb 6 '17 at 21:15