# How to find a series for comparison with $\sum 1/\sqrt{n(n+1)}$?

The series $$\sum_{n=1}^{+\infty} \frac{1}{\sqrt{n(n+1)}}$$ I have tried ratio and integral both lead me to inconclusive, so probably it's by comparisson but I can't find What to compare.

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The integral test is inconclusive? What did you get for $$\int_1^{\infty}{dx\over\sqrt{x(x+1)}}$$ –  Gerry Myerson Aug 13 '14 at 13:22

Try a Limit Comparison with $\sum \frac{1}{n}$.
Along similar lines, note that $n(n+1)\lt n(4n)$, so $\dfrac{1}{\sqrt{n(n+1)}}\gt \dfrac{1}{2n}$.
The informal idea is that for large $n$, the number $\sqrt{n(n+1)}$ is roughly equal to $n$.
Yes, for the limit comparison you then divide top and bottom by $n$. You will get $\frac{1}{\sqrt{1+\frac{1}{n}}}$, and the limit is $1$. My second suggestion is more "concrete" so you may prefer it. –  André Nicolas Aug 13 '14 at 13:34
So for the limit comparison the algebraic "trick" one uses is $\sqrt{n(n+1)}=n\sqrt{1+\frac{1}{n}}$. –  André Nicolas Aug 13 '14 at 13:47