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How do I find an upper bound for the partial sums $\sum_{k=0}^{n} \frac1{k!}$ using a geometric progression?

I tried saying that $\sum_{k=0}^{n} \frac1{k!} < \sum_{k=0}^{\infty} \frac1{2^k}$ for large $n$ but this doesn't really give a valid upper bound: $\frac{1}{1-\frac12}=2$ because of the few exceptions that arise with small $n$. I understand that I could change the starting index so it is such that the inequality is true for all $n$ and then I could add the removed terms onto 2 to get a valid upper bound but I was wondering if there was a cleaner way to do this?

Thanks!

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    $\begingroup$ $$e = \sum_{k \geq 0} \frac{1}{k!}$$ $\endgroup$ – MathematicsStudent1122 Dec 20 '17 at 5:11
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    $\begingroup$ Yes this is what is trying to be proved ... $\endgroup$ – Aakash Lakshmanan Dec 20 '17 at 5:15
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    $\begingroup$ @AakashLakshmanan it is the definition of $e$ $\endgroup$ – ℋolo Dec 20 '17 at 5:49
  • $\begingroup$ @holo well historically it arises as the limit of $(1+\frac 1n)^n$, the definition using power series came next. $\endgroup$ – zwim Dec 20 '17 at 6:00
  • $\begingroup$ @zwim it is irrelevant $\endgroup$ – ℋolo Dec 20 '17 at 6:05
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You can also simply shift the inequality : $k!\ge 2^{k-1}$ is now true for all $k\ge 0$.

And you get $\displaystyle \sum\limits_{k=0}^{\infty}\dfrac 1{k!}\le 2\sum\limits_{k=0}^{\infty} \dfrac 1{2^k}\le 4$

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  • $\begingroup$ Wow, that is pretty. Thank you! $\endgroup$ – Aakash Lakshmanan Dec 20 '17 at 8:19
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The main idea is right, but you have to be more precise. The inequality $k!\geq 2^k$ is wrong for $k\in\{1,2,3,4\}$ but true for all $k\geq 5$ (Proof by induction!). So you get $$ \sum_{k=0}^\infty\frac1{k!}=\sum_{k=0}^4\frac1{k!}+\sum_{k=5}^\infty\frac1{k!}=\frac{65}{24}+\sum_{k=5}^\infty\frac1{k!}\leq \frac{65}{24}+\sum_{k=5}^\infty\frac1{2^k}. $$

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  • $\begingroup$ Ya this is what I had in mind ... so there is no pretty way to do this? $\endgroup$ – Aakash Lakshmanan Dec 20 '17 at 5:21
  • $\begingroup$ No. The convergence of a series does not depend on the starting index. You can say, that $\sum_{k=5}^\infty \frac1{k!}$ converges since $\sum_{k=5}^\infty\frac1{2^k}$ converges and add the missing terms on both sides of the inequality. $\endgroup$ – Mundron Schmidt Dec 20 '17 at 5:22
  • $\begingroup$ If you are willing to compute the first five terms, you can use the inequality $k!\ge 6^{k-5}5!$ to get a tighter upper bound for $\sum_{k=5}^\infty\frac1{k!}$. $\endgroup$ – Lord Shark the Unknown Dec 20 '17 at 5:36

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