Absolute convergence tests Hi I am interested in the following series:
$$\sum_{n=1}^{\infty}(-1)^{n}(\sqrt{n+3}-\sqrt{n})$$
I have been able to show that this series converges by proving the Leibnitz test. Does anyone know how to show that it converges absolutely? What tests are available for this?
Thanks for assistance.
 A: Notice that
$$\sqrt{n+3}-\sqrt n\ge \sqrt{n+1}-\sqrt n$$
and obviously the series
$$\sum_{n\ge1}\sqrt{n+1}-\sqrt n$$
is divergent by telescoping. The given series isn't absolutely convergent.
A: *

*Convergence.
You may write 
$$
\begin{align}
\sum_{n=4}^{\infty}(-1)^{n}(\sqrt{n+3}-\sqrt{n})
&=\sum_{n=4}^{\infty}(-1)^{n}\sqrt{n}\left(\sqrt{1+\frac 3n}-1\right)\\\\
&=\sum_{n=4}^{\infty}(-1)^{n}\sqrt{n}\left(1+\frac{3}{2n}-\frac{9}{8n^2}+\mathcal{O}\left(\frac{1}{n^2}\right)-1\right)\\\\
&=\frac{3}{2}\sum_{n=4}^{\infty}\frac{(-1)^{n}}{\sqrt{n}}-\frac{9}{8}\sum_{n=4}^{\infty}\frac{(-1)^{n}}{n\sqrt{n}}+\sum_{n=4}^{\infty}\mathcal{O}\left(\frac{1}{n\sqrt{n}}\right)\\\\
\end{align}
$$ and your initial series converges, being the sum of convergent series.


*

*Absolute convergence.
The same reasoning gives, for $N$ great,
$$
\begin{align}
\sum_{n=4}^{N}(\sqrt{n+3}-\sqrt{n})&=\sum_{n=4}^{N}\sqrt{n}\left(\sqrt{1+\frac 3n}-1\right)\\\\
&=\sum_{n=4}^{N}\sqrt{n}\left(1+\frac{3}{2n}-\frac{9}{8n^2}+\mathcal{O}\left(\frac{1}{n^2}\right)-1\right)\\\\
&=\frac{3}{2}\sum_{n=4}^{N}\frac{1}{\sqrt{n}}-\frac{9}{8}\sum_{n=4}^{N}\frac{1}{n\sqrt{n}}+\sum_{n=4}^{N}\mathcal{O}\left(\frac{1}{n\sqrt{n}}\right)
\end{align}
$$ and the series of absolute terms, being the sum of a divergent series and convergent series, is convergent and your initial series is absolutely divergent.
A: By the definition of absolutely convergent series you have to prove that $\sum_{n=1}^\infty|(-1)^n(\sqrt{n+3}-\sqrt{n})|$ doesn't converge
$$S_k=\sum_{n=1}^k |(-1)^n(\sqrt{n+3}-\sqrt{n})|=\sum_{n=1}^k\sqrt{n+3}-\sqrt{n}=\sqrt{k+3}-1$$
$$\lim_{k\to\infty}S_k=\infty$$
From this we can say that the series isn't aboslutely convergent
