How to show $\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$ $$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$
Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this. 
WolframAlpha says it is true but I don't know what the gamma function is.
$$ \sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}$$
So then I need to show that$$ \sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big]$$
Is it possible to do this by induction? I don't really know how to approach this now.
 A: It suffices to show that $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \frac{1}{n!}$$
For all $k \ge n+1$, we have $$k! \ge 2^{k-n} \cdot n!$$
Therefore, we have $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \sum_{k=n+1}^\infty \frac{1}{n! \cdot 2^{k-n}} = \frac{1}{n!} \sum_{i=1}^\infty \frac{1}{2^i} = \frac{1}{n!}$$ 
A: Another way: 
$$
\sum_{k=n}^{\infty} \frac{1}{k!} = \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots \bigg) < \frac{1}{n!} \bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots \bigg)\\
= \frac{1}{n!} \cdot \frac{n+1}{n} = \frac{1}{n!} \cdot \bigg(1 + \frac{1}{n} \bigg)
$$
Now compare this expression to what you have on the RHS: certain terms cancel out. What do you get?  
A: $$\dfrac{1}{(k+1)!} = \dfrac{1}{(k-1)!}(\dfrac{1}{k} - \dfrac{1}{k+1}) \leq \dfrac{1}{(n-1)!}(\dfrac{1}{k}-\dfrac{1}{k+1})$$ when $k\geq n$
So
$$\sum_{k=n}^\infty\dfrac{1}{(k+1)!} \leq \dfrac{1}{(n-1)!}\sum_{k=n}^\infty (\dfrac{1}{k}-\dfrac{1}{k+1}) = \dfrac{1}{(n-1)!} *\frac{1}{n} = \dfrac{1}{n!}$$
