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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? –  Did Apr 7 '11 at 8:19

3 Answers 3

If $n\ge3$, $\frac12(n+1)<n$ and $n\ge\mathrm{e}$ hence $n^{n-1}\ge\mathrm{e}^{n-1}$ and it is enough to prove that $n<n^{3/2}$, 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??? $ $ –  Did Mar 27 '11 at 2:28

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