Using Jensen's inequality, prove: $$~(1) ~~n!<n^n<(n!)^2~~~~~ and ~~~(2)~~~ e^n\geqslant \frac{n^n}{n!}.$$

Attempt. For (1), since $\log$ is strictly concave, using Jensen: $$\log\frac{1+2+\ldots+n}{n}>\frac{\log1+\log2+\ldots+\log n}{n}$$ equivalently $$\log\frac{\frac{n(n+1)}{2}}{n}>\frac{\log n!}{n}$$ equivalently $$\left(\frac{n+1}{2}\right)^n> n!,$$ and since $\frac{n+1}{2}<n$ (for $n>1$), we get the first part. How can one get the second part?

Is $(2)$ a consequence of $(1)$ (regarding somehow that $\left(1+\frac{1}{n}\right)^n<e$)?

Thank in advance.

  • $\begingroup$ Yep I've mispelled it $\endgroup$ – JoseSquare Feb 18 at 12:06
  • $\begingroup$ As an aside, if $n\in \mathbb{N}$, then (2) follows immediately from the Taylor series expansion of $e^n$ ($e^n =\sum_{k\ge 0} \frac{n^k}{k!}$). $\endgroup$ – Minus One-Twelfth Feb 18 at 13:03
  • $\begingroup$ Also, note that $(2!)^2 = 2^2$, so the first inequality holds only for $n > 2$. $\endgroup$ – астон вілла олоф мэллбэрг Feb 18 at 14:08

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