# Factorial sum estimate $\sum_{n=m+1}^\infty \frac{1}{n!} \le \frac{1}{m\cdot m!}$

Prove that:

$$\displaystyle \sum_{n=m+1}^\infty \dfrac{1}{n!} \le \dfrac{1}{m\cdot m!}$$

I have tried induction on $m$ but it does not work very well. Any suggestion?

$$\sum_{n=m+1}^{\infty} \frac{1}{n!}=\frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)(m+2)}+\frac{1}{(m+1)(m+2)(m+3)}+\ldots \right)$$ $$\leq \frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)^2}+\frac{1}{(m+1)^3}+\ldots \right)=\frac{1}{m!}\frac{1}{m+1}\frac{1}{1-\frac{1}{m+1}}=$$ $$\frac{1}{m!}\frac{1}{m+1-1}=\frac{1}{m!}\frac{1}{m}$$
hint: $\dfrac{1}{(m+1)!}+\dfrac{1}{(m+2)!}+\cdots=\dfrac{1}{m!}\left(\dfrac{1}{m+1}+\dfrac{1}{(m+1)(m+2)}+\cdots\right)\le \dfrac{1}{m!}\left(\dfrac{1}{m+1}+\dfrac{1}{(m+1)^2}+ \dfrac{1}{(m+1)^3}\cdots \right)...$