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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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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Feb 23 '13 at 17:06
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 – Mike Feb 23 '13 at 19:12
up vote 1 down vote accepted

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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