Convergence and absolute convergence of $\sum\limits_{n=1}^{\infty}(-1)^n\frac{\cos\sqrt{n+1}-\cos\sqrt{n}}{n}$ $$\sum_{n=1}^{\infty}(-1)^n\frac{\cos\sqrt{n+1}-\cos\sqrt{n}}{n}.$$
For absolute convergence I've tried using formula for difference of cosines and then wanted to use comparison test, but don't really know how to estimate $$\sum_{n=1}^{\infty}\left|\frac{-2\sin(\frac{\sqrt{n+1}-\sqrt{n}}{2})\sin(\frac{\sqrt{n+1}+\sqrt{n}}{2})}{n}\right|.$$
 A: By MVT
$$|\cos(\sqrt{n+1})-\cos(\sqrt{n})|$$
$$=(\sqrt{n+1}-\sqrt{n})|-\sin(c_n)|$$
$$\leq \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
$$\leq \frac{1}{2\sqrt{n}}$$
$$\implies |u_n|\leq \frac{1}{2n^{\frac{3}{2}}}$$
$\implies \sum u_n$ absolutly convergent.$(\frac{3}{2}>1)$.
A: This is expanding on Did's outline, from the comments.

Starting from where you left, we wish to bound
$$
a_n\stackrel{\rm def}{=}\left\lvert \frac{2\sin\left(\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)\sin\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)}{n}\right\rvert
$$
Using the facts that


*

*$\left\lvert \sin\frac{\sqrt{n+1}+\sqrt{n}}{2}\right\rvert \leq 1$
and

*$\left\lvert \sin\frac{\sqrt{n+1}-\sqrt{n}}{2}\right\rvert \leq \frac{\sqrt{n+1}-\sqrt{n}}{2}$ (from $\left\lvert\sin x\right\rvert \leq \left\lvert x\right\rvert$ for every $x\in\mathbb{R}$)


we get
$$
0\leq a_n \leq \frac{\sqrt{n+1}-\sqrt{n}}{n}
$$
and it only remains to conclude by comparison with a $p$-series, recalling that (for instance) $$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \leq \frac{1}{2\sqrt{n}}$$
