Determine the convergence of $\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{n(n-1)}$ So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test.
$$
\sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)}
$$
You could do a limit comparison with $\dfrac{1}{n^{3/2}}$

Or we could be a bit witty. Write $\displaystyle \sum_{n = 2} \frac{\sqrt{n +1}}{n(n-1)} = \sum_{n = 1} \frac{\sqrt{n + 2}}{(n+1)n} &lt; \sum_{n = 1} \frac{\sqrt{3n}}{n^2} = \sqrt 3 \sum_{n=1} \frac{1}{n^{3/2}}$

So that's just basic comparison, no limits or anything.
$\dfrac{\sqrt{n+1}}{n(n-1)}\le\dfrac{{n+1}}{n-1}=(1+\dfrac{1}{n})(1-\dfrac{1}{n})^{-1}=(1+\dfrac{1}{n})(1+\dfrac{1}{n}-\cdots)=1+\dfrac{2}{n}+O(\dfrac{1}{n^2}).$

$\displaystyle\sum\dfrac{{n+1}}{n-1}$ is convergent and so is the given series.