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?
Proving $\frac{1}{2}(n+1)<\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$ without induction
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real-analysis
inequality
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1Did you get something out of one of the answers below? – 2011-04-07
3 Answers
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If $n\ge3$, $\frac12(n+1)
0
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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3And??? $ $ – 2011-03-27