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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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    Is this a question about rings, groups, $\Bbb Z$-modules, $\Bbb Q$ vector spaces? – 2017-01-04
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    @user2520938 Sorry if this isn't tagged correctly. I couldn't decide where this would be an appropriate question to ask. I just wanted to know how would I show these two sets equal, that's all. – 2017-01-04
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    If $n>b$ then $n!=b\times m$ so $\frac ab=\frac {am}{n!}$. – 2017-01-04
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    @user264885 see my answer to hopefully understand the relevance of my comment. – 2017-01-04
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    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. – 2017-01-04
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    @lulu Yes your interpretation is correct and that is the question I was asking. – 2017-01-04
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    So, then my argument shows this. It doesn't show uniqueness, but of course there isn't any trivial sort of uniqueness, as $\frac 46 = 4\times \frac 1{3!}=\frac 1{2!}+\frac 1{3!}$ for example. Maybe it's like [Egyptian fractions](http://mathworld.wolfram.com/EgyptianFraction.html) in that the greedy algorithm generates a viable expression and you can call that the representation. Not sure if that works out well. – 2017-01-04
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    Say you want $\frac{a}{b},\ a,b > 0$, then $\frac{a}{b}= a(b-1)! \cdot \frac{1}{b!}$. – 2017-01-04
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    Thanks to everyone for all the help. – 2017-01-04
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    You could avoid all of these problems about not knowing whether you were talking about groups or rings by writing $({\mathbb Q},+)$. – 2017-01-04

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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.