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The series is: $\sum_{n=1}^\infty {(\frac{n}{n+2})}^{n^2}$. I'm relatively certain it converges, but I'm not sure how to prove this. It was suggested that I try and use the comparison test, but I'm not sure which series I could use to compare it with.

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WA says that it uses the ratio test. wolframalpha.com/input/… –  Sabyasachi Mar 15 at 20:18

3 Answers 3

up vote 3 down vote accepted

$(1+\frac{2}{n})^n\ge 1+\frac{2}{n}.n=3$

It follows that, $(\frac{n}{2+n})^{n^2}\le \frac{1}{3^n}$

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I'm a little confused how you derived the first line : $(1+\frac{2}{n})^n\ge 1+\frac{2}{n}.n$ –  userzzz23 Mar 15 at 20:25
    
@Kyle its the binomial theorem $(1+x)^n \ge 1+ {n\choose 1}x + {n \choose 2}x^2 + .. \ge 1+nx$ –  r9m Mar 15 at 20:27
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Even easier, @Kyle: it is Bernoulli's Inequality.en.wikipedia.org/wiki/Bernoulli's_inequality –  DonAntonio Mar 15 at 20:28

Hints:

$$a_n:=\left(\frac n{n+2}\right)^{n^2}=\frac1{\left(1+\frac2n\right)^{n^2}}\implies$$

$$\sqrt[n]{a_n}=\frac1{\left(1+\frac2n\right)^n}\xrightarrow[n\to\infty]{}e^{-2}$$

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Apparently someone didn't like my answer....oh, well. –  DonAntonio Mar 15 at 20:23
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Personally I like it.+1 –  Sami Ben Romdhane Mar 15 at 20:36

Using Taylor series we prove easily the convergence of the series by comparison:

$$\left(\frac{n}{n+2}\right)^{n^2}=\left(1+\frac2n\right)^{-n^2}=\exp\left(-n^2\log\left(1+\frac2n\right)\right)\sim_\infty e^{-2n}=_\infty o\left(\frac1{n^2}\right)$$

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