I saw proofs of $e < (1+\frac{1}{n})^{n+0.5}$, but how to prove the inequality:
$ e > (1+\frac{1}{n})^{n+\frac{1}{k}} $
for $k>2$ and sufficiently large $n$? I'm interested in elementary solutions (without using derivatives or Taylor expansion), suitable for early stages of a calculus course.
Cauchy-Schwarz says $$ \begin{align} \log\left(1+\frac1n\right) &=\int_n^{n+1}\frac1x\,\mathrm{d}x\tag{1a}\\ &\le\left(\int_n^{n+1}\frac1{x^2}\,\mathrm{d}x\right)^{1/2}\left(\int_n^{n+1}1\,\mathrm{d}x\right)^{1/2}\tag{1b}\\ &=\frac1{\sqrt{n(n+1)}}\tag{1c} \end{align} $$ That is, $$ \left(1+\frac1n\right)^{\sqrt{n(n+1)}}\le e\tag2 $$ For any $\epsilon\gt0$, if $n\ge\frac1{8\epsilon}+\frac\epsilon2-\frac12$, then $$ \begin{align} \left(n+\frac12-\epsilon\right)^2 &=n(n+1)+\overbrace{\frac14+\epsilon^2-2\left(n+\frac12\right)\epsilon}^{\le0}\tag{3a}\\ &\le n(n+1)\tag{3b} \end{align} $$ Therefore, putting together $(2)$ and $(3)$, we get that for $n\ge\frac1{8\epsilon}+\frac\epsilon2-\frac12$ $$ \left(1+\frac1n\right)^{n+\frac12-\epsilon}\le e\tag4 $$
