# Alternating series of reciprocals of factorials

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## Solution 1

(UPDATED)
Concerning the finite sum : $$\tag{1}S_n:=\sum_{k=2}^n\frac{(-1)^k}{k!}$$ we have simply : $$\tag{2} S_n=\frac 1{n!}\left[\frac{n!}e\right]=\frac{!n}{n!}$$ with $[x]$ the nearest integer (i.e. the round function) and $\,!n\,$ the number of derangements for $n$ elements (added from Wood's answer (+1) in the related thread for additional properties).

To prove this directly you may use a method similar to the one proposed in this thread.

The idea is that the remaining terms of the Maclaurin expansion of $\,e^{-1}\;n!\,$ (after the $n$ first terms) are : $\;\displaystyle \frac {(-1)^{n+1}}{(n+1)}+ \frac {(-1)^{n+1}}{(n+1)(n+2)} +\cdots\;\$ and thus with absolute value bounded by $\dfrac 12$ for $n>1$ that will disappear with the round function.

## Solution 2

Take the hint that $$\sum_{n=0}^{\infty}\frac {x^n}{n!} =e^x$$ What can you conclude from this? Hope it helps.

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Consider the sum: $$\frac{(-1)^2}{2!}+\frac{(-1)^3}{3!}+\frac{(-1)^4}{4!}+\ldots+ \frac{(-1)^n}{n!}.$$ Does there exist a nice closed form of such sum?
Compare to $e^x$.