calculus - Proving $sumlimits_n^infty frac{1}{sqrt{n(n+1)}}$ divergent without integral test

Evaluate if the following series is convergent or divergent: $\sum\limits_n^\infty\frac 1 {\sqrt{n(n+1)}}$.

I could use the integral test that would prove me this series to be divergent. However I want to prove them divergent using Weierstrass comparison theorem.

$$\sum_n^\infty \frac{1}{\sqrt{n(n+1)}}>\sum_n^\infty \frac 1 {n(n+1)} = \sum_n^\infty \frac{1}{n^2+n}=\text{?}$$

However I cannot find a series that are smaller than the last.
I tried to find any inequality to bring $n^2$ down to $n$, but I was not successful.

Question:

How can I find a smaller divergent series for $\sum_\limits{n}^{\infty}\frac{1}{{n^2+n}}$?

Thanks in advance!

Answer

Since $n(n+1)<4n^2$ for every natural $n$, you have $$(\forall n\in\mathbb{N}):\frac1{\sqrt{n(n+1)}}>\frac1{2n}.$$