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Does the sum $\displaystyle \sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^{n^2}$ converge?

I was given a hint to use the comparison test but fail to find a suitable series for comparison. I am not looking for a full solution but simply a convergent larger sequence for comparison.

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MathMajor
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2 Answers2

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Given series is Convergent.

Hint : Use Cauchy's root test.

Let , $a_n=\left(\frac{n}{n+1}\right)^{n^2}$. Then , $a_n^{1/n}\to 1/e<1$.

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  • Great, thanks! This is what I was looking for. – MathMajor Nov 06 '15 at 10:53
  • Since you seem to be great at series, any chance you could look at my second and last question? http://math.stackexchange.com/questions/1515874/convergence-of-sum-n-1-infty-frac1-log-nnp-for-p-ge-0 – MathMajor Nov 06 '15 at 11:01
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Considering, for $n$ sufficiently large, $$A_n=\left(\frac{n}{n+1}\right)^{n^2}$$ $$ \log(A_n)=-n^2\log(1+\frac 1n)=-n+\frac{1}{2}-\frac{1}{3 n}+\frac{1}{4 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ $$A_n=\left(\frac{n}{n+1}\right)^{n^2}=\exp \left(-n+\frac{1}{2}-\frac{1}{3 n}+O\left(\left(\frac{1}{n}\right)^2\right)\right)<e^{\frac 12 -n}$$ This last inequality is valid for any $n \geq 0$ and then, as a final result $$\sum_{n=1}^\infty A_n=\sum_{n=1}^\infty (\frac{n}{n+1})^{n^2}<\sum_{n=1}^\infty e^{\frac 12 -n}=\frac{\sqrt{e}}{e-1}$$

Claude Leibovici
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