<s>[INST] 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.
 [/INST] The left hand side is
$$
\begin{align}
&\left|\,\left(n+\frac12\right)\log\left(1+\frac1n\right)-1\,\right|\\
&=\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\frac1{4n^4}+\dots\right)-1\\
&=\frac1{3\cdot4n^2}-\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}-\frac4{6\cdot10n^5}+\dots+\frac{(-1)^k(k-1)}{2k(k+1)n^k}+\dots\tag{1}
\end{align}
$$
The right hand side is
$$
\begin{align}
&\left|\,\left(n-\frac12\right)\log\left(1-\frac1n\right)-1\,\right|\\
&=\left(n-\frac12\right)\left(\frac1n+\frac1{2n^2}+\frac1{3n^3}+\frac1{4n^4}+\dots\right)+1\\
&=2+\frac1{3\cdot4n^2}+\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}+\frac4{6\cdot10n^5}+\dots+\frac{(k-1)}{2k(k+1)n^k}+\dots\tag{2}
\end{align}
$$
Thus, the left hand side tends to $0$ and the right hand side tends to $2$.

However, assuming that the inequality is actually
$$
\left|\,\log\left(\left(1+\frac1n\right)^{n+\frac12}\cdot\frac1e\right)\,\right| \le\left|\,\log\left(\left(1-\frac1n\right)^{n-\frac12}\cdot e\right)\,\right|\tag{3}
$$
The right hand side is
$$
\begin{align}
&\left|\,\left(n-\frac12\right)\log\left(1-\frac1n\right)+1\,\right|\\
&=\left(n-\frac12\right)\left(\frac1n+\frac1{2n^2}+\frac1{3n^3}+\frac1{4n^4}+\dots\right)-1\\
&=\frac1{3\cdot4n^2}+\frac2{4\cdot6n^3}+\frac3{5\cdot8n^4}+\frac4{6\cdot10n^5}+\dots+\frac{(k-1)}{2k(k+1)n^k}+\dots\tag{4}
\end{align}
$$
Thus, the left hand side is still smaller than the right hand side, but the difference is much smaller.
</s>