Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the sum of this series?

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

share|improve this question
3  
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

3 Answers 3

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.

share|improve this answer
    
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 $x=-1$, $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{n!} = 0$$

share|improve this answer

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.