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

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.

  • 5
    $\begingroup$ Just in case: the inequality doesn't hold for $n = 0$. For $n \geqslant 1$, note that $(n+k)! \geqslant n!\cdot (n+1)^k$. $\endgroup$ – Daniel Fischer Dec 3 '15 at 10:44
  • $\begingroup$ $$\sum_{k=n}^{\infty}{\frac{1}{k!}}\le\frac{2}{n!}\Longleftrightarrow$$ $$e-\frac{e\Gamma(n,1)}{\Gamma(n)}\le\frac{2}{n!}\space\text{for}\space n>-1$$ $\endgroup$ – Jan Dec 3 '15 at 10:51

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

  • $\begingroup$ Does the inequality $k! \geq 2^{k-n} \cdot n!$ have a name? I think I need to prove that in order to use it. $\endgroup$ – elfeck Dec 3 '15 at 11:06
  • $\begingroup$ Not really. For a proof note that $\frac{k!}{n!} = k(k-1) \cdots (n+1) \ge 2 \cdot 2 \cdots 2 = 2^{n-k}$. Of course, this doesn't hold when $n=0$, so you need to exclude that. $\endgroup$ – Gyumin Roh Dec 3 '15 at 11:06
  • $\begingroup$ Okay thanks. Since in my case I substitute $n$ for $n+1$ anyway $n=0$ does not cause issues. Thanks for your help! $\endgroup$ – elfeck Dec 3 '15 at 11:10

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?

  • $\begingroup$ Small typo, that $k$ should start from $n$, not $n+1$. $\endgroup$ – Gyumin Roh Dec 3 '15 at 11:18
  • $\begingroup$ yeah, that's how I solved it anyway $\endgroup$ – Alex Dec 3 '15 at 11:44

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


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


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