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Show$$(n-1)!<\frac{n^ne}{e^n}<n!$$ for $n\geq 2$.

My attempt: use induction and natural log $\ln$.

i) $n=2$: $(n-1)!=1!=1<\frac{2^2e}{e^2}=\frac{4}{e}\simeq 1.48<2=2!=n!$

ii) assume the inequality holds for $n$.

iii) apply $\ln$: $\ln(n-1)!=\sum_{k=1}^{n-1}\ln k<\ln\frac{n^ne}{e^n}=n\ln n+1-n<\ln n!=\sum_{k=1}^{n}\ln k$. that is, $$\sum_{k=1}^{n-1}\ln k<n\ln n+1-n<\sum_{k=1}^{n}\ln k$$

and from this inequality, I want to deduce $$\sum_{k=1}^{n}\ln k<(n+1)\ln (n+1)+1-(n+1)<\sum_{k=1}^{n+1}\ln k$$

for the left inequality, $$\sum_{k=1}^{n-1}\ln k+\ln n<n\ln n+1-n+\ln n=(n+1)\ln n+1-n<(n+1)\ln n+1-n$$

so I have to get $(n+1)\ln n+1-n < (n+1)\ln(n+1)-n$, that is, $$1<(n+1)\ln\frac{n+1}{n}$$ or $$e<(\frac{n+1}{n})^{n+1}$$

but $(\frac{n+1}{n})^{n+1}=(1+\frac{1}{n})^{n+1}\to 1$ as $n\to\infty$, so there exists $N>0$ s.t. $e>(\frac{n+1}{n})^{n+1}$ for all $n>N$.

So, I don't know what to do from here. I guess the induction may not be applicable. any hints?

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  • $\begingroup$ Your inequality can be killed by the Stirling's formula. $\endgroup$ – Michael Rozenberg Jul 9 '17 at 7:22
  • $\begingroup$ Yep it seems asymptotically wrong. Did you try to flank with factorials instead of having the term $n!$ flanked like in this thread : math.stackexchange.com/questions/23814/… ? Maybe adding the $\sqrt{2\pi n}$ term somewhere can re-establish the inequality, but I guess the purpose was not too. $\endgroup$ – zwim Jul 9 '17 at 7:37
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You forgot that $e = \lim_{n\to\infty}(1+\frac{1}{n})^n.$ In general, $$(1+\frac{1}{n})^n<e<(1+\frac{1}{n})^{n+1}$$ holds for all positive integers $n$.

This means your last assumption is wrong: $$(1+\frac{1}{n})^N\to 1$$, for any constant $N$, but that is not the case here, because we have $N = n+1.$

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  • $\begingroup$ Thank you, my problem was in fact equivalent to $(1+1/n)^n<e<(1+1/n)^{n+1}$. $\endgroup$ – user159234 Jul 9 '17 at 7:44

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