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Determine whether the following series converges: $$\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2}.$$

I tried to solve it using ratio's test but it leads me nowhere.

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Hint (without the number $e$). Note that for $n\geq 1$, by the binomial theorem, $$\left(\frac{n+1}{n}\right)^n=\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}\geq 1+\frac{n}{n}=2\implies \left(\frac{n}{n+1}\right)^{n^2}\leq \frac{1}{2^n}.$$

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Hint: $\left(\frac n{n+1}\right)^n = \left(1- \frac 1{n+1}\right)^n\to \frac 1e$. So there is a bound $N$ such that for any $n>N$ the corresponding term of the sum will be smaller than, for instance, $\dfrac1{2^n}$

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Note that

$$\left(\frac{n}{n+1}\right)^{n^2}=\left(\frac{1}{\left(1+\frac1n\right)^n}\right)^{n}\sim \frac1{e^n}$$

thus the series converges by comparison with $\sum \frac1{e^n}$.

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    $\begingroup$ You can't compare to $\sum\frac1{e^n}$, because the terms of the sequence are larger than $\frac1{e^n}$. However, any number smaller than $e$ you can use. $\endgroup$ – Arthur Jan 24 '18 at 9:19
  • $\begingroup$ @Arthur I mean by comparison test as limit $$\frac{\left(\frac{n}{n+1}\right)^{n^2}}{\frac1{e^n}}\to l\in \mathbb{R}$$ $\endgroup$ – gimusi Jan 24 '18 at 9:30
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The ratio test works like a charm $$a_n=\left(\frac{n}{n+1}\right)^{n^2}\implies \log(a_n)=n^2 \log \left(\frac{n}{n+1}\right)=n^2 \log \left(1-\frac{1}{n+1}\right)$$ Using Taylor series $$\log(a_n)=n^2\left(-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)\right)=-n+\frac{1}{2}-\frac{1}{3 n}+O\left(\frac{1}{n^2}\right)$$ $$\log(a_{n+1})-\log(a_n)=-1+\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{ a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{e}+\frac{1}{3 e n^2}+O\left(\frac{1}{n^3}\right)$$

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