I am trying to show $(1-\frac{1}{2n})\displaystyle\sum_{k=0}^{n}\frac{1}{k!}\leq \left(1+\frac{1}{n}\right)^n\forall\,n\in\mathbb{N}\backslash\{1\}$
There is a hint $(1-\gamma_{1})\cdots (1-\gamma_{m})\geq 1-(\gamma_{1}+\cdots+\gamma_{m})$ where $\gamma_i\in(0,1)$
I have tried to find a tight lower bound so that I might reach the desires inequality.
Here's my attempt:
$\displaystyle\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cdot\frac{1}{n^k}=\sum_{k=0}^n\frac{1}{k!}\left(\frac{n-1}{n}\right)\cdots\left(\frac{n-k+1}{n}\right)=\sum_{k=0}^n\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\geq\sum_{k=0}^n\frac{1}{k!}\left(1-\left(\frac{1}{n}+\cdots+\frac{k-1}{n}\right)\right)=\sum_{k=0}^n\frac{1}{k!}\left(1-\frac{k(k-1)}{2n}\right)$
However, I think the bound that I got above is less than the desired one.
I cannot think of other way to use the hint better. Might I get some clue here?
Well, you were close: \begin{align}\sum_{k=0}^n\frac{1}{k!}\left(1-\frac{k(k-1)}{2n}\right)&=\sum_{k=0}^n\frac{1}{k!}-\frac1{2n}\sum_{k=0}^n\frac{k(k-1)}{k!}\\ &=\sum_{k=0}^n\frac{1}{k!}-\frac1{2n}\sum_{k=2}^n\frac{1}{(k-2)!}\\ &=\sum_{k=0}^n\frac{1}{k!}-\frac1{2n}\sum_{k=0}^{n-2}\frac{1}{k!}\\ &\ge\left(1-\frac1{2n}\right)\sum_{k=0}^n\frac{1}{k!} \end{align}