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How to prove the inequality $${(n+1)\over e^n}^n<n!$$

I have tried mathematical induction, but it doesn't work! Are there other methods to solve it?

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You can try and bring $n!$ on the left side and $e^n$ on the right and then use the definition of $e$(the $(1+\frac{1}{n})^n$ definition).Not sure if it will work though. – rah4927 Apr 20 '14 at 7:14
Thank you! I got it – python3 Apr 20 '14 at 7:20
I don't know if my method works.If you could tell us what you have gotten,it will help us. – rah4927 Apr 20 '14 at 7:25
Stirling approximation of $n!$ could help. – Claude Leibovici Apr 20 '14 at 7:33
If you find an answer ok can you please mark it as such? Thanks. – Umberto Apr 20 '14 at 7:50
up vote 9 down vote accepted

Use the binomial theorem on $(1 + n)^n$ and try.

$$(1 + n)^n = 1 + \frac{n!}{1!(n - 1)!} n + \frac{n!}{2!(n - 2)!} n^2 + \frac{n!}{3!(n - 3)!} n^3 + \dots + \frac{n!}{n!(n - n)!} n^n \\ = n! \{\frac{1}{0! n!} + \frac{n}{1! (n-1)!} + \frac{n^2}{2! (n-2)!} + \frac{n^3}{3! (n-3)!} + \dots + \frac{n^n}{n! (n-n)!}\} \\ \le n! \{1 + \frac{n}{1!} + \frac{n^2}{2!} + \frac{n^3}{3!} + \dots\} \\ = n! e^n$$

Hope it is clear now.

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Let $$ a_n=\frac{(n+1)^n}{n!} $$ then $$ \begin{align} a_n &=\left(1+\frac1n\right)^n\frac{n^n}{n!}\\ &=\left(1+\frac1n\right)^n\frac{n^{n-1}}{(n-1)!}\\ &=\left(1+\frac1n\right)^n\,a_{n-1}\\[9pt] &\le ea_{n-1} \end{align} $$ Since $a_1=2\lt e$, inductively, we have that $a_n\lt e^n$.

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To prove that inequality, it means that we need to prove $${(n+1)^n\over n!}<e^n$$

Because${(1+{1\over n})^n}<e$, so ${(n+1)^n\over n!}$=${(1+{1\over n})^n}$${(1+{1\over n-1})^{n-1}}$``````$2^1\over1$$1\over1!$$<e^n$

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Stirling approximation write $$n! \simeq \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ Then the inequality is satisfied if $$n+1 \lt (2 \pi )^{\frac{1}{2 n}} n^{\frac{1}{2 n}+1} \simeq (2 \pi )^{\frac{1}{2 n}}[n+\frac{1}{2} \log \left(n\right)]$$ which is satisfied for any $n \geq 1$

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