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The containment of one side is obvious, but I can't see how to show $\mathbb{Q} \subseteq\langle \frac{1}{n!}: n\in \mathbb{N} \rangle$, and would prefer an answer that shows the set containment via algebraic techniques.

I am mean how to show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)?

That is, I am considering the rationals as an additive group and $\langle \cdot \rangle$ means the subgroup generated by the set.

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    $\begingroup$ Is this a question about rings, groups, $\Bbb Z$-modules, $\Bbb Q$ vector spaces? $\endgroup$ – user2520938 Jan 4 '17 at 10:46
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    $\begingroup$ If $n>b$ then $n!=b\times m$ so $\frac ab=\frac {am}{n!}$. $\endgroup$ – lulu Jan 4 '17 at 10:49
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    $\begingroup$ Note: I am assuming that you meant "show that every rational number can be written as a finite sum of the reciprocals of factorials (or their negatives)" . If you meant something else, you should clarify. $\endgroup$ – lulu Jan 4 '17 at 10:50
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    $\begingroup$ @lulu Yes your interpretation is correct and that is the question I was asking. $\endgroup$ – Mark Jan 4 '17 at 10:52
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    $\begingroup$ Say you want $\frac{a}{b},\ a,b > 0$, then $\frac{a}{b}= a(b-1)! \cdot \frac{1}{b!}$. $\endgroup$ – Zubzub Jan 4 '17 at 11:02
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It follows from the Fundamental Theorem of Arithmetic that $\Bbb Q$ is generated, as a group, by the set $$ \left\{\frac1{p^k}\text{ such that $p$ is prime and $k>0$}\right\} $$ (one can make this set of generators smaller, but that is irrelevant here). Then it is enough to show that each $1/p^k$ can be obtained from the $1/n!$, but then it is enough to notice that $$ \frac1{p^k}=(p^k-1)!\frac1{(p^k)!}. $$

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HINT: The generated subgroup also contains any finite sum of the same element $k$ times since $$\frac{k}{n!} = \underbrace{\frac{1}{n!} + \ldots \frac{1}{n!}}_{k \text{ times}}.$$

EDIT: I hope I understood the question correctly. If not, please leave a comment.

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