Show divergence of infinite series $\sum \frac{1}{\sqrt{(n(n+1)}}$ I am having trouble showing that this series is divergent. I do see that
$\frac{1}{\sqrt{(n(n+1)}} = \frac{1}{n\sqrt{1+ \frac{1}{n}}}$.
However, I can't find a series that is smaller and diverges to infinity, as $n \leq n\sqrt{1+ \frac{1}{n}}$. 
I have tried the root test as well, but I am not sure wether the following is correct:
$\lim_{n \to \infty} \frac{1^{\frac{1}{n}}}{n^{\frac{1}{n}}((1+ \frac{1}{n})^{\frac{1}{n}})^{\frac{1}{2}}} = \frac{1}{1} = 1$.
Edit:
$\sqrt(n)\sqrt(n+1) = n\sqrt{1+\frac{1}{n}} \leq n \sqrt{2} < 2n$ for all $n \in \mathbb{N}$, thus follows
$0 \leq \frac{1}{2n} \leq \frac{1}{\sqrt{n(n+1)}}$.
$\frac{1}{2} \sum \frac{1}{n}  = \sum \frac{1}{2n}$ diverges, so by the comparison test $\sum \frac{1}{\sqrt{(n(n+1)}}$ also diverges.

$$
\frac 1 {\sqrt{n(n+1)}} \ge \frac 1 {\sqrt{(n+1)^2}} = \frac 1 {n+1} \text{ and } \sum_{n=1}^\infty \frac 1 {n+1} =\infty.
$$
PS inspired by a comment below:
\begin{align}
\sum_{n=1}^\infty \frac 1 {n+1} & = \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots \\[10pt]
& = \left( 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 \cdots \right) - 1 \\[10pt]
& = \left( \sum_{n=1}^\infty \frac 1 n \right) - 1.
\end{align}
If you change only finitely many terms of a sequence (in this case by deleting only one term), you don't alter whether the series converges or not, nor whether the divergence is to $\infty$.
