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. $$\sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)}$$

• Did you try a limit comparison test with $\displaystyle\sum \frac1{n^{3/2}}$? Your numerator is roughly $n^{1/2}$ and your denominator is roughly $n^2$, so your fraction is roughly $1/n^{3/2}$. – Brian M. Scott Apr 9 '12 at 5:30
• @BrianM.Scott: the fraction is less than $(n-1)^{-3/2}$ for $n\ge2$. – robjohn Jan 9 '15 at 16:21

You could do a limit comparison with $\dfrac{1}{n^{3/2}}$

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

So that's just basic comparison, no limits or anything.

If $a_n\sim b_n (n\rightarrow\infty)$, then $\sum\limits_n a_n$ and $\sum\limits_n b_n$ converge together or diverge together. $\frac{\sqrt{n+1}}{n(n-1)}\sim \frac{\sqrt{n}}{n^{2}}=\frac{1}{n^{3/2}}$, and $\sum\limits_n \frac{1}{n^{3/2}}$ is a convergent $p$-series.

• That's sometimes called a $p$-series, but the term harmonic series is reserved for $\sum_n \frac{1}{n}$ – Robert Israel Apr 9 '12 at 7:40
• Yes you're right, I apologize for my inaccuracy! – Mark Apr 9 '12 at 7:49
• At least one of the two sequences need to be nonnegative for this to be true (that $a_n \sim b_n$ implies they converge or diverge together). Take $\sum {(-1)^n \over \sqrt{n}}$, a convergent series, and $\sum {(-1)^n \over \sqrt{n} + (-1)^n}$. – Najib Idrissi Apr 9 '12 at 7:50

Creative telescoping is another chance. Since, for any $n\geq 1$: $$\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}=\frac{2}{(\sqrt{n+1}+\sqrt{n})\sqrt{n(n+1)}}\geq\frac{\sqrt{n+3}}{(n+1)(n+2)}\tag{1}$$ (it can be proved through induction) we have, for any $N\geq 3$:

$$\begin{eqnarray*}\sum_{n=2}^{N}\frac{\sqrt{n+1}}{n(n-1)}&=&\sum_{n=1}^{N-1}\frac{\sqrt{n+2}}{n(n+1)}=\frac{\sqrt{3}}{2}+\sum_{n=1}^{N-2}\frac{\sqrt{n+3}}{(n+1)(n+2)}\\&\leq&\frac{\sqrt{3}}{2}+2\sum_{n=1}^{N-2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\\&\leq&\frac{\sqrt{3}}{2}+2.\tag{2}\end{eqnarray*}$$

$\dfrac{\sqrt{n+1}}{n(n-1)}\le\dfrac{{n+1}}{n-1}=(1+\dfrac{1}{n})(1-\dfrac{1}{n})^{-1}=(1+\dfrac{1}{n})(1+\dfrac{1}{n}-\cdots)=1+\dfrac{2}{n}+O(\dfrac{1}{n^2}).$

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

• How can a series be convergent if its general term does not tend to zero? – Jack D'Aurizio Jan 9 '15 at 16:01