# 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.

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I edited your latex to make it appear as what I think it should. (moved brackets and backslashes around) Can you double check that the inequality is correctly stated? –  Calvin Lin Feb 23 '13 at 17:53
@julien, you can't multiply by $e$, because the logarithm turns products into sums, which may in turn yield sign changes, and because absolute values are taken, you can't just remove them on both sides. –  Lieven Feb 23 '13 at 18:02
@Lieven Sure. Either I was blind, or these parentheses have changed in between. Checked. Yeah, they've been changed... –  1015 Feb 23 '13 at 18:04
You can use limits –  Example Mo Feb 23 '13 at 19:12

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*}

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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.

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