how could we show that the following holds for $k \ge 2$, $k \in \mathbb{N}$: $ 1 < \left( k - \frac{1}{2}\right) \log{\left(\frac{k}{k-1}\right)} $ or equivalently, how can we show that $ e < \left( \frac{k}{k-1} \right)^{k-1/2} $ Thanks!
inequality proof
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calculus
real-analysis
inequality
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0@joriki Mine was incorrect. – 2012-09-27
3 Answers
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For $k\ge2$,
$ \begin{align} \left(k-\frac12\right)\log\frac k{k-1} &= -\left(k-\frac12\right)\log\frac{k-1}k \\ &= -\left(k-\frac12\right)\log\left(1-\frac1k\right) \\ &\gt \left(k-\frac12\right)\left(\frac1k+\frac1{2k^2}+\frac1{3k^3}\right) \\ &= 1+\frac1{12k^2}-\frac1{6k^3} \\ &\ge1\;. \end{align} $
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0@Apin: You're welcome! – 2012-09-27
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An article on this problem appeared in the American Mathematical Monthly 2 years ago. It contains 3 proofs. The reference is:
Amer. Math. Monthly 117 (2010) 273–277. doi:10.4169/000298910X480126
You can view the article here.
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You can try this.
1- Let k-1=u,
2- Show the function $(1+\frac{1}{u})^{u+\frac{1}{2}}$ is decreasing for $u\geq 1$,
3- Show the limit of this function when u goes to infinity is $e$,
4- Conclude.