TITLE: Proving $(1 + 1/n)^{n+1} \gt e$
QUESTION [14 upvotes]: I'm trying to prove that
$$ \left(1 + \frac{1}{n}\right)^{n+1} > e $$
It seems that the definition of $e$ is going to be important here but I can't work out what to do with the limit in the resulting inequality.

REPLY [6 votes]: I would like to post a short proof of a stronger claim, namely that:

$$ \forall n\geq 1,\qquad \left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\geq e.\tag{1} $$

By switching to logarithms, $(1)$ is equivalent to:
$$ \left(n+\frac{1}{2}\right)\int_{-1/2}^{1/2}\frac{dx}{n+\frac{1}{2}+x}\,dx \geq 1 \tag{2}$$
but $g(x)=\frac{1}{x+1}$ is a convex function on $\mathbb{R}^+$, hence $(2)$ trivially follows from the Hermite-Hadamard inequality.