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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.

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

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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

3 Answers 3

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

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

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

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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.

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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
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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

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

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

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