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?
Alternating series of reciprocals of factorials
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$\begingroup$
summation
factorial
closed-form
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0Compare to $e^x$. – 2017-01-05
3 Answers
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Given series is exponential series, so you have
$\sum_{k=0}^{\infty}\frac {x^k}{k!} =e^x $
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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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(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.