How to prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent? 
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.
 A: 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.
A: 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.
