How to show $\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$

Question

$$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$

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.

Answer 1

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 $$\sum_{k=n+1}^\infty \frac{1}{k!} \le \sum_{k=n+1}^\infty \frac{1}{n! \cdot 2^{k-n}} = \frac{1}{n!} \sum_{i=1}^\infty \frac{1}{2^i} = \frac{1}{n!}$$

Answer 2

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?

Answer 3

$$\dfrac{1}{(k+1)!} = \dfrac{1}{(k-1)!}(\dfrac{1}{k} - \dfrac{1}{k+1}) \leq \dfrac{1}{(n-1)!}(\dfrac{1}{k}-\dfrac{1}{k+1})$$ when $k\geq n$

So

$$\sum_{k=n}^\infty\dfrac{1}{(k+1)!} \leq \dfrac{1}{(n-1)!}\sum_{k=n}^\infty (\dfrac{1}{k}-\dfrac{1}{k+1}) = \dfrac{1}{(n-1)!} *\frac{1}{n} = \dfrac{1}{n!}$$