# Prove this following inequality

Show that, for all integers m > 1, $\frac {1}{2me}$ < $\frac {1}{e}$ - $(1-\frac{1}{m})^m$ < $\frac {1}{me}$

Here's one part: We have $\left(1+\frac1{km}\right)^{km}\to e$ as $k\to\infty$, hence for $k$ big enough the error in $$\left(1-\frac1m\right)^m\cdot e\approx\left(1-\frac1m\right)^m\left(1+\frac1{km}\right)^{km}= \left[\left(1-\frac1m\right)\left(1+\frac1{km}\right)^k\right]^m$$ becomes arbitrarily small. Using $(1+x)^n> 1+nx$ if $x>0,n\ge2$ (Bernoulli) twice, we get $$\left[\left(1-\frac1m\right)\left(1+\frac1{km}\right)^k\right]^m> \left[\left(1-\frac1m\right)\left(1+\frac1{m}\right)\right]^m=\left(1-\frac1{m^2}\right)^m>1-\frac1m.$$ Thus $\left(1-\frac1m\right)^m\cdot e>1-\frac1m$, i.e. $$\frac1e-\left(1-\frac1m\right)^m<\frac1{em}.$$