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I want to show that $\displaystyle\frac{1}{2}(n+1)<\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$. But except induction, I do not know how I could prove this?

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    Did you get something out of one of the answers below?2011-04-07

3 Answers 3

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If $n\ge3$, $\frac12(n+1) and $n\ge\mathrm{e}$ hence $n^{n-1}\ge\mathrm{e}^{n-1}$ and it is enough to prove that $n, which is obviously true. Then you can check manually the cases $n=1$ (for which the strict inequality is false, by the way) and $n=2$.

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That looks a lot like Stirling's approximation for the factorial. You are asking to show $\frac{n+1}{2} \lt \frac {n! e}{\sqrt{2\pi}}$ in that approximation, and $n!$ grows very fast. So you can use Stirling for large $n$, perhaps supplemented with specific calculations for small $n$.

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This is equivalent to showing $n \leq 2 \frac{n^{n+\frac{1}{2}}}{e^{n-1}}-1 = 2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-1 $

Let $f(n) = 2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-n-1$

Then $f(n) \geq 0$ (i.e. find critical points and use first derivative test).

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    And??? $ $2011-03-27