# Prove that $\sum_{k=0}^n\frac{1}{k!}\geq \left(1+\frac{1}{n}\right)^n$ [duplicate]

It basically says it all in the title. I tried solving the inequality using the bernoulli inequality somehow $$\dfrac{\displaystyle\sum_{k=0}^n\frac{1}{k!}}{(1+\frac{1}{n})^n}\geq 1,$$ but the factorial gets me everytime.

Also I've tried induction without success:

$\displaystyle\sum_{k=0}^{n+1}\frac{1}{k!}\stackrel{IV}\geq \left(1+\frac{1}{n}\right)^n+\frac{1}{(n+1)!}$

## marked as duplicate by Sasha, Clayton, Prahlad Vaidyanathan, Emily, Sujaan KunalanNov 26 '13 at 16:54

Expand the right-hand side using the Binomial Theorem. We get a sum of $n+1$ terms. The generic term is $$\binom{n}{k}\frac{1}{n^k}.$$ This is $$\frac{1}{k!} \frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}.$$ But it is clear that $\frac{n(n-1)\cdots (n-k+1}{n^k}\le 1$, since in the numerator we have a product of $k$ terms, each $\le n$.
Binomial theorem $$(1+\frac1n)^n=1+1+\frac1{2!}(1-\frac1n)+\dotsb+\frac1{k!}(1-\frac1n)(1-\frac2n)\dotsm(1-\frac{k-1}n)+\dotsm$$ now
$$\frac1{k!}(1-\frac1n)(1-\frac2n)\dotsm(1-\frac{k-1}n)\le\frac1{k!}$$
$$\sum_{k=0}^n\frac{1}{k!}\geq (1+\frac{1}{n})^n$$