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Consider the sequence $\{x_n \}$ defined by $$x_n = e \left (\frac {n} {n+1} \right )^{n + \frac 1 2},\ n \geq 1.$$

Prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent.

My attempt: I find that \begin{align*} \sum\limits_{n=1}^{\infty} \log x_n & = \sum\limits_{n=1}^{\infty} \left ( 1 - \left (n + \frac 1 2 \right ) \log \left (1 + \frac 1 n \right ) \right ) \\ & = - \sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \left ( \frac {1} {m+1} - \frac {1} {2m} \right ) \frac {1} {n^m} \\ & = -\sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \frac {m-1} {2m(m+1)} \frac {1} {n^m} \end{align*}

From here how do I proceed further? Please help me in this regard.

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  • $\begingroup$ Show that the summands of your first expression divided by $n^{-2}$ have finite limit as $n\to\infty$. Then use the convergence of the sum of $n^{-2}$. $\endgroup$ – vujazzman Apr 8 '20 at 7:02
  • $\begingroup$ The integral test works fine $\endgroup$ – Claude Leibovici Apr 8 '20 at 7:15
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Use Taylor series to second and third order: $$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)=1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\dots\right)$$ $$\le1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}\right)=\frac1{4n^2}$$ $$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)\ge1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}\right)=-\frac{n+2}{12n^3}$$ Thus $|\log x_n|\le\frac1{4n^2}$. Since $\sum_{n=1}^\infty\frac1{4n^2}$ converges, $\sum_{n=1}^\infty\log x_n$ converges.

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Notice, that:

$$\sum_{i=1}^n \log x_i= \log \prod_{i=1}^n x_i = \log \left( e\left(\frac{1}{2}\right)^{\frac{3}{2}} e\left(\frac{2}{3}\right)^{\frac{5}{2}} \dots e\left(\frac{n}{n+1}\right)^{\frac{2n+1}{2}} \right)= \log \left( e^n \frac{n!}{(n+1)^{n+\frac{1}{2}}}\right)$$

Using Stirling's approximation:

$$\lim_{n \to \infty} \log \left( e^n \frac{n!}{(n+1)^{n+\frac{1}{2}}}\right) = \lim_{n \to \infty} \log \left( e^{n + 1} \frac{(n+1)!}{(n+1)^{n + 1} \sqrt{n+1}} \frac{1}{e}\right) = \log \frac{\sqrt{2 \pi}}{e}$$

Note, that this is the exact value of the sum, not only the proof of convergence.

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    $\begingroup$ This solution is nice as it actually yields the sum. $\endgroup$ – vujazzman Apr 8 '20 at 7:13
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    $\begingroup$ @vujazzman yes, I wanted to boast a bit, but forgot;> Thank you, sir! $\endgroup$ – Andronicus Apr 8 '20 at 7:14

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