It seems that $\sum_{n=1}^\infty \left(\frac 1 {\sqrt {n+1}} - \frac 1 {\sqrt {n+2}}\right)$ converges but I can't prove it

Question

When I write out a few terms in the series it is clear that the sum is telescoping towards $\frac 1 {\sqrt 2}$ $\frac 1 {\sqrt 2} - \frac 1 {\sqrt 3} + \frac 1 {\sqrt 3} - \frac 1 {\sqrt 4} + \frac 1 {\sqrt 4} ...$

It is known that $\sum_{n=1}^\infty \frac 1 {\sqrt n}$ diverges, but I can't use that with the direct comparison test since it is always greater than $\frac 1 {\sqrt {n+1}}$ and likewise $\frac 1 {\sqrt {n+2}}$ although this wouldn't help me anyway.

I tried the nth term test but I got $0$, and for the ratio test if I try just with $\frac 1 {\sqrt {n+1}}$ I get $\lim_{n \to \infty} \frac 1 {\sqrt {n+2}} \cdot \frac {\sqrt {n+1}} 1 = 1$

This is inconclusive. Maybe if I can prove that $\frac 1 {\sqrt {n+1}}$ and $\frac 1 {\sqrt {n+2}}$ both diverge similarly so that when they are subtracted they cancel eachother out?

Answer 1

The partial sum is $s_{n}=\displaystyle\sum_{k=1}^{n}\dfrac{1}{\sqrt{k+1}}-\dfrac{1}{\sqrt{k+2}}=\dfrac{1}{\sqrt{1+1}}-\dfrac{1}{\sqrt{n+2}}\rightarrow\dfrac{1}{\sqrt{2}}$ as $n\rightarrow\infty$, so the series converges.

It is the telescoping which finishes the job.

Answer 2

Note that

$$\frac 1 {\sqrt {n+1}} - \frac 1 {\sqrt {n+2}}\sim\frac12\frac{1}{n\sqrt n}$$