Absolute convergence of the series $\sum_1^{\infty} {(-1)}^n(\sqrt{n+1} -\sqrt{n})$ $\sum_1^{\infty} {(-1)}^n(\sqrt{n+1} -\sqrt{n})$
I want to know if this series diverge or if it is absolutely convergent or conditionally convergent. I used Leibniz' Criterion of alternating series and I think that  it converges conditionally, but I'm not sure if it is absolutely convergent.
Can someone help me out please?
 A: Hint:
The absolute value of the general term has an equivalent:
$$\sqrt{n+1}-\sqrt n=\frac 1{\sqrt{n+1}+\sqrt n}\sim_\infty \frac1{2\sqrt n}. $$
Does the latter converge?
A: $$
\sqrt{n+1}-\sqrt{n}=\sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right)
$$
And
$$
\sqrt{1+\frac{1}{n}}\underset{(+\infty)}{=}1+\frac{1}{2n}+o\left(\frac{1}{n}\right)
$$
So
$$
\sqrt{n+1}-\sqrt{n}\underset{(+\infty)}{=}\frac{1}{2\sqrt{n}}+o\left(\frac{1}{\sqrt{n}}\right)
$$
The series $\displaystyle \sum \frac{1}{\sqrt{n}}$ diverges, so it does not converge absolutely.
A: $\begin{array}\\
\text{If}\\
s(m)
&=\sum_{n=1}^{m} {(-1)}^n(\sqrt{n+1} -\sqrt{n})\\
\text{then}\\
s(2m)
&=\sum_{n=1}^{2m} {(-1)}^n(\sqrt{n+1} -\sqrt{n})\\
&=\sum_{n=1}^m (-(\sqrt{2n}-\sqrt{2n-1})+(\sqrt{2n+1}-\sqrt{2n}))\\
&=\sum_{n=1}^m (\dfrac{-1}{\sqrt{2n}+\sqrt{2n-1}}+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n}})\\
&=\sum_{n=1}^m \dfrac{-(\sqrt{2n+1}+\sqrt{2n})+\sqrt{2n}+\sqrt{2n-1}}{(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\
&=\sum_{n=1}^m \dfrac{-\sqrt{2n+1}+\sqrt{2n-1}}{(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\
&=\sum_{n=1}^m \dfrac{-2}{(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})}\\
\text{so}\\
s(2m)
&>-2\sum_{n=1}^m \dfrac{1}{8(2n-1)^{3/2}}\\
&=-\frac14\sum_{n=1}^m \dfrac{1}{(2n-1)^{3/2}}\\
\text{and}\\
s(2m)
&<-2\sum_{n=1}^m \dfrac{1}{8(2n+1)^{3/2}}\\
&=-\frac14\sum_{n=1}^m \dfrac{1}{(2n+1)^{3/2}}\\
\end{array}
$
and both bounds converge.
Therefore the
whole sum converges since
$(\sqrt{n+1}-\sqrt{n})
=\dfrac1{\sqrt{n+1}+\sqrt{n}} \to 0$.
Note:
Wolfy says that
the sum of the first 1,000,000 terms
is about -0.23954,
validates my algebra,
and says that
$(\sqrt{2n+1}+\sqrt{2n-1})(\sqrt{2n}+\sqrt{2n-1})(\sqrt{2n+1}+\sqrt{2n})\\
=8(2n)^{3/2}
-\dfrac{5}{2\sqrt{2n}}
-\dfrac{17}{32(2n)^{5/2}}
+O(n^{-9/2})
$.
