Error bounds for $e$

Question

Prove that for all $n\in\mathbb{N}_+$, we have $$(1+\frac{1}{n})^n>\sum_{k=0}^n \frac{1}{k!}-\frac{e}{2n}.$$

Answer 1

Note that when $x>0$, we have: $$\log(1+x)>x-\frac{x^2}{2},\quad e^{-x}>1-x.$$ It follows that $$(1+\frac{1}{n})^n=e^{n\log(1+\frac{1}{n})}>e^{n(\frac{1}{n}-\frac{1}{2n^2})}=e\cdot e^{-\frac{1}{2n}}>e(1-\frac{1}{2n})>\sum_{k=0}^n\frac{1}{k!}-\frac{e}{2n}.$$