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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!

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@joriki Mine was incorrect. –  Julián Aguirre Sep 27 '12 at 12:26

3 Answers 3

up vote 6 down vote accepted

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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Quite interestingly, this version of the "usual" formulation of this problem (where $k$ replaced by $k+1$) leads to your simpler solution as the signs in the log series don't alternate any more. Presented with the original problem I was unable to think of this shift so I've now learned a neat trick. –  Ragib Zaman Sep 27 '12 at 12:41
Thanks joriki!! –  Apin Grann Sep 27 '12 at 12:50
@Apin: You're welcome! –  joriki Sep 27 '12 at 12:56

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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Thank you Ragib!! –  Apin Grann Sep 27 '12 at 12:50

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.

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Thanks, Kaye!!! –  Apin Grann Sep 27 '12 at 12:51
You're welcome! –  Tomás Sep 29 '12 at 16:10

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