I have an interesting series whose sum I know will be $0$, but not able to prove it. The series is

$$\sum_{i=0}^r \frac{(-1)^{i}}{i! (r-i)!}=0.$$ I had checked it for several values of $\text {“}r\text{''}$ and got the result, but was not able to show it. I got this observation while proving the Leibniz formula in case of distributions.

Any type of help will be appreciated. Thanks in advance.

  • 1
    $\begingroup$ Binomially expand $(1-1)^r$. $\endgroup$ Commented Aug 26, 2017 at 19:55

2 Answers 2


Because its just $r!(1-1)^r=0$

  • $\begingroup$ what if i say that $\text r\in \mathbf R^n$ $\endgroup$
    – bunny
    Commented Aug 26, 2017 at 21:08
  • $\begingroup$ @bunny That doesn't even make sense. $\endgroup$ Commented Aug 27, 2017 at 1:07
  • $\begingroup$ @bunny If $r\in\mathbb R^n$ then let $n=2$ and $r=(3,2)$. So we need to prove that $\sum\limits_{i=0}^{(3,2)}\frac{(-1)^i}{i!((3,2)-i)!}=0.$ What is it??? $\endgroup$ Commented Aug 27, 2017 at 7:30

Multiply your sum by $r!$ You will obtain :

$$\sum_{i=0}^r \binom{r}{i} (-1)^{i}(1)^{n-i}$$

which is $0$ because it is the binomial expansion of $(1-1)^r.$


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