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

Using Jensen's inequality, prove: $$~(1) ~~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 $$\frac{n+1}{2} (for $$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)?

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