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What is the sum of this series?

$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$

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Do you know of any power series which you may be able to leverage? Any time I see factorial denominators, I think about Taylor series... – process91 Aug 23 '12 at 16:08
up vote 10 down vote accepted

Hint: We have $$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots.$$ Multiply both sides by $x$ and differentiate.

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Doh!!!!!! I can't believe I didn't see it! Thanks. :) – Legendre Aug 23 '12 at 16:15

Alternatively, write it as:

$$1-\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!}... +\\ \left(-\frac{1}{1!}+ \frac{2}{2!} - \frac{3}{3!}...\right)$$

The first line is $e^{-1}$ and the second line, after cancelling terms, you see is $-e^{-1}$

More generally, if $$(z)_i = z(z-1)...(z-(i-1))$$ is the falling factorial, and $p(z) = a_0(z)_0 + a_1(a)_1 + ... a_k(z)_k$, then:

$$\sum_{n=0}^\infty \frac{p(n)}{n!} x^n = e^x (a_0 + a_1x + a_2x^2 + ... a_k x^k)$$

In this case, $p(z) = 1 + z = (z)_0 + (z)_1$ so

$$\sum_{n=0}^\infty \frac{n+1}{n!} x^n = e^x (1+x)$$

And, in particular, for $x=-1$, $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{n!} = 0$$

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Maybe one can do it without using power series: $$ \begin{align} \sum_{n=0}^{\infty}(-1)^n\frac{n+1}{n!} &=\sum_{n=0}^{\infty}(-1)^n\frac{n}{n!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\ &=\sum_{n=1}^{\infty}(-1)^n\frac{1}{(n-1)!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\ &=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{1}{k!}+\sum_{n=0}^{\infty}(-1)^n\frac{1}{n!}\\&=0. \end{align} $$

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while this is nice (+1)... I would have taken one more step explicitely for a complete housekeeping... – Gottfried Helms Oct 12 '12 at 21:05

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