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I need to determine if the following sum is convergent or divergent.$$\sum_{n=1}^\infty \bigl(\frac{n}{n+2}\bigr)^{n^2}$$

I proceeded using the simplified root test but i'm stuck here $$\lim_{n\to\infty}\bigl(\frac{n}{n+2}\bigr)^n$$ I considered using the euler factorisation but it just doesn't work out.

Danxe
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  • In the first expression, there is an exponent $n^2$ but it became $n$ in the second. It is normal ? – Claude Leibovici Mar 20 '16 at 14:58
  • @ClaudeLeibovici I already applied the root test but i didn't explicitly right the intermediary steps. that's the reason for it. – Danxe Mar 20 '16 at 15:33

1 Answers1

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$$\left(\dfrac n{n+2}\right)^n=\left(1+\dfrac2n\right)^{-n} =\left[\left(1+\dfrac2n\right)^{n/2}\right]^{-2}$$

Now $\lim_{n\to\infty}\left(1+\dfrac2n\right)^{n/2}=e$

lab bhattacharjee
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