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.
Best Answer
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.