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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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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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