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I need to determine whether the following series converges or diverges:

$$ \sum_{n=1}^\infty \frac{n}{(1+n)\left( 1 + n \sqrt{n}\right)} $$

My first idea was to show that $a_n$ tends to zero, and so it does. My next plan was to rewrite the sum as

$$ a_n = \frac{1}{1+n\sqrt{n}} - \frac{1}{(n+1)\left( 1 + n\sqrt{n} \right)} $$

Where the firt one can be compared to a p-series, and the second one is smaller than the first one so it also converges. Is this correct? Is there an easier way to show that the series converge?

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You can compare it directly to $\sum \frac{1}{n\sqrt{n}}$ –  N. S. May 2 '12 at 12:27
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Do you need to evaluate it, as you claim, or just decide whether or not it converges? –  TonyK May 2 '12 at 12:27
    
I just need to decide whether it converges or diverges. Sorry on behalf of my non-existent English capabilities. –  N3buchadnezzar May 2 '12 at 12:53
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@N3buchadnezzar: To answer your question, your solution is good, but relies a bit too much on the simplicity of the expression. A direct size estimate is the way to go, you should say to yourself that the $1$'s at the bottom don't matter in the long run (actually, they help, but that's irrelevant). So in the long run your terms are effectively $\frac{n}{n^2\sqrt{n}}$. Then make a limit comparison with $p$-series, or a direct comparison since the $1$'s help. –  André Nicolas May 2 '12 at 14:34
    
Title modified. –  Did May 2 '12 at 15:51

3 Answers 3

up vote 6 down vote accepted

$$\frac{n}{(1+n)(1+n\sqrt{n})} \leq \frac{n}{n \cdot n \cdot \sqrt{n}}=\frac{1}{n^{3/2}}.$$

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A wise and simple move! –  Chris's sis Aug 17 '12 at 14:08
    
Thank you very much! :-) –  Siminore Aug 17 '12 at 15:21

For deciding convergence, Siminore's comparison test answers affirmatively. If for real $x>0$ we wish to approximate the sum over $\mathbb{Z}^+$ of $$ \eqalign{ f(x) &=\frac{x}{(x+1)(x^{3/2}+1)} =\left(1-\frac1{1+x}\right)\left(1+x^{3/2}\right)^{-1}\\ &=\frac1{1+x^{3/2}} -\frac1{\left(1+x\right)\left(1+x^{3/2}\right)} =\frac1{\left(1+x^{-1}\right)\left(1+x^{3/2}\right)} } $$ then its antiderivative is $$ F(x) = \arctan\left(\sqrt{x}\right) + \frac16 \, \log \frac{ \left(x - \sqrt{x} + 1\right)^4 }{ \left(x + 1\right)^3 \left(\sqrt{x} + 1\right)^2 } $$ and a rough lower bound on the sum woud be $$ S = \sum_{x=1}^\infty f(x) > I = \int_1^\infty f(x)\,dx = \frac{\pi}4 + \frac56 \, \log2 \approx 1.363021 $$ since $f(x)$ is strictly decreasing for $x > 0.73173541$

f(x) as a continuous function in blue, and as a sequence in red

since its global maximum is at the root $t\approx0.85541534$ of the denominator $3t^5+t^3-2$ (for $t=\sqrt x$) of $$ f'(x)= -\frac{ 3x^{5/2}+x^{3/2}-2 }{ 2\,\left(x+1\right)^2 \left(x^3+2x^{3/2}+1\right)} \,. $$ We can think of our sum $S$ as a left-endpoint (improper) Riemann sum. A correspondingly rough upper bound would be $f(1)+I=\frac14+I\approx1.613021$. A better approximation would be the average of these two: $$ S \approx \frac18 + \frac{\pi}4 + \frac56 \, \log2 \approx 1.488021 \,. $$ The antiderivative $F$ above can be derived by setting $x=t^2$ (so $dx=2t\,dt$) and $$ g(t)=f(t^2)=\frac{t^2}{(t^2+1)(t^3+1)} $$ so that $$ \eqalign{ t\,g(t)& =\tfrac13\frac{2t-1}{t^2-t+1} -\tfrac12\frac{t-1}{t^2+1} -\tfrac16\frac1{t+1} \\& =\tfrac13\frac{2t-1}{t^2-t+1} -\tfrac14\frac{2t}{t^2+1} -\tfrac16\frac1{t+1} +\tfrac14\frac1{t^2+1} } $$ and, for $t>0$ (to dispense with absolute values inside the natural logarithm), $$ \eqalign{ F(x)& = \int f(x)\,dx =2\int t\,g(t)\,dt \\& = \frac23\ln\left(t^2-t+1\right) - \frac12\ln\left(t^2 +1\right) - \frac13\ln\left(t +1\right) + \frac12\arctan{t} } $$ which yields the above in $x$.

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What do you mean by “integral approximation” ? Is 6 an integral approximation of $\pi^2$ ? –  Lierre May 2 '12 at 13:04
    
@Lierre He means an approximation of the infinite series by the associated integral. $$\sum f(x) \sim \int f(x) dx$$ –  Pedro Tamaroff May 2 '12 at 20:41
    
@Lierre: thanks, my wording was not good. I think I fixed it now. –  bgins May 2 '12 at 22:09
    
@PeterTamaroff & bgins — What I meant is that I think it is inaccurate to say that the sum of the series is approximated by the associated integral. What would mean $\pi^2 \approx \int_1^\infty x^{-2} dx = 6$ ? However, I strongly agree with the details that bgins added : comparaison with integral can lead to upper and lower bounds of the total sum, and the value of the integral can be refined. In good cases, it could be refined in a real sequence of approximations with Euler-McLaurin formula. –  Lierre May 3 '12 at 8:08
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@Lierre: yes, you are of course right. It's a crude approximation, or a rough bound. I fixed the wording to avoid any confusion. –  bgins May 3 '12 at 9:23

If two sequences (nonnegative) $a_{n}\sim b_{n} (n\rightarrow\infty)$ then $\sum_{n}a_{n},\sum_{n}b_{n}$ converge together or diverge together. $\frac{n}{(1+n)(1+n\sqrt{n})}\sim \frac{1}{n\sqrt{n}}=\frac{1}{n^{3/2}}$ converges.

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