Proof of inequality involving logarithms How could we show that 
$$\left|\log\left( \left({1 + \frac{1}{n}}\right)^{n + \frac{1}{2}}\cdot \frac{1}{e}\right)\right| \leq \left|\log\left( \left({1 - \frac{1}{n}}\right)^{n - \frac{1}{2}}\cdot \frac{1}{e}\right)\right| ,\; \forall n \text{ sufficiently large?} $$
I already calculated in wolfram the limit of the quotient of the logs when $n \rightarrow \infty$. And it is zero. However, I can't prove it.

It is enough to prove that
$$\lim_{n\rightarrow \infty }\log \left( \left( 1+\frac{1}{n}\right) ^{n+1/2}
\frac{1}{e}\right) =0$$
and
$$\lim_{n\rightarrow \infty }\log \left( \left( 1-\frac{1}{n}\right) ^{n-1/2}\frac{1}{e}\right) =-2.$$
The first limit can be evaluated as follows:
$$\begin{eqnarray*}
\lim_{n\rightarrow \infty }\log \left( \left( 1+\frac{1}{n}\right) ^{n+1/2}\frac{1}{e}\right)  &=&\lim_{n\rightarrow \infty }\left(n+\frac{1}{2}\right)\log \left( 1+\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }n\log \left( 1+\frac{1}{n}\right)+\frac{1}{2}\lim_{n\rightarrow \infty }\log \left( 1+\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }\frac{\log \left( 1+\frac{1}{n}\right) }{\frac{1}{n}}+0-1 \\
&=&1-1=0;
\end{eqnarray*}$$
and the second:
\begin{eqnarray*}
\lim_{n\rightarrow \infty }\log \left( \left( 1-\frac{1}{n}\right) ^{n-1/2}\frac{1}{e}\right)  &=&\lim_{n\rightarrow \infty }\left(n-\frac{1}{2}\right)\log \left( 1-\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }n\log \left( 1-\frac{1}{n}\right) -\frac{1}{2}%
\lim_{n\rightarrow \infty }\log \left( 1-\frac{1}{n}\right) -1 \\
&=&\lim_{n\rightarrow \infty }\frac{\log \left( 1-\frac{1}{n}\right) }{\frac{1}{n}}-0-1 \\
&=&-1-1=-2.
\end{eqnarray*}
