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I have the following series as an exercise but for some reason i cannot prove if it converges or not.

I used the integral test. The series is positive and decreasing so we can use it( i will not write the proof here). That's what the theory says.

The series is the following

$\sum\limits_{n=1}^{\infty }\frac{1}{n \sqrt{n+1}}$

The result i calculate is the following

$2\lim\limits_{t\rightarrow \infty }\frac{1}{3}\sqrt{(t+1)^{3}}-\sqrt{t+1}-\frac{1}{3}\sqrt{1+1}+\sqrt{1+1}= $

$2\lim\limits_{t\rightarrow \infty }\frac{1}{3}\sqrt{(t+1)^{3}}-\sqrt{t+1}+\frac{2}{3}\sqrt{2}$

Can someone help?

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How about comparison to $n^{-3/2}$? –  Gerry Myerson Oct 29 '11 at 12:03
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2 Answers

up vote 3 down vote accepted

Note that, $0<\frac{1}{n\sqrt{n+1}}<\frac{1}{n^{3/2}}$. Since $$ \lim_{N\rightarrow\infty}\int_{n=1}^N \frac{1}{n^{3/2}} \mathrm{d} n = \lim_{N\rightarrow\infty} 2-\frac{2}{\sqrt{N}} = 2 <\infty , $$ we find that $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$ is convergent by the integral test. Therefore $\sum_{n=1}^\infty \frac{1}{n\sqrt{n+1}}$ is convergent by the comparison test.

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How did we decide to make a comparison with this. –  pleis_j Oct 29 '11 at 12:09
    
@pleis: well, $\frac1{n\sqrt{n}}$ looks awfully like the original term, so it makes sense to start comparing with that... –  J. M. Oct 29 '11 at 12:13
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For every $n\geqslant1$, $$\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}=\frac1{\sqrt{n}\sqrt{n+1}(\sqrt{n+1}+\sqrt{n})}, $$ and $\sqrt{n}\leqslant\sqrt{n+1}\leqslant\sqrt2\sqrt{n}$ hence $$ \frac1{1+\sqrt2}\frac1{n\sqrt{n+1}}\leqslant\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}\leqslant\frac12\frac1{n\sqrt{n+1}}. $$ This telescoping series has sum $$ \sum\limits_{n=1}^{+\infty}\left(\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}\right)=1, $$ hence $$ 2\leqslant\sum\limits_{n=1}^{+\infty}\frac1{n\sqrt{n+1}}\leqslant1+\sqrt2, $$ in particular this series with nonnegative terms converges.

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