Proving $\mathrm e <3$ Well I am just asking myself if there's a more elegant way of proving $$2<\exp(1)=\mathrm e<3$$ than doing it by induction and using the fact of  $\lim\limits_{n\rightarrow\infty}\left(1+\frac1n\right)^n=\mathrm e$, is there one (or some) alternative way(s)? 

The sequence $x_n=(1+1/n)^{n+1}$ is strictly decreasing and converges to $e.$ So $e<x_5=2.985984 <3.$ 
Remark: $\ln x_n=-(n+1)\ln (1-1/(n+1))=1+\sum_{j=1}^{\infty}(j+1)^{-1}(n+1)^{-j}.$ Comparing the terms of this series to the corresponding terms  in the series for $\ln x_{n+1} , $ we see that $\ln x_n>\ln x_{n+1} , $ so $ x_n>x_{n+1}.$ 
