# Value of convergent infinite factorial series [duplicate]

I've tried searching for this both with "infinite factorial series" and on approach0.xyz to no avail.

Wolfram|Alpha gives: $$\sum_{i=0}^{\infty}\frac{i!}{(i+4)!}=\frac{1}{18}$$ I was wondering where this came from. It's the infinite limit of the partial sum formula given by Wolfram|Alpha, as expected, but that also raises the question of where the partial sum formula comes from.

Any help would be appreciated.

• Sep 19, 2019 at 16:26

Noting $$\frac{1}{(x+1) (x+2) (x+3) (x+4)}=-\frac12\bigg(\frac{1}{x+2}-\frac{1}{x+3}\bigg)+\frac{1}{6}\bigg(\frac1{x+1}-\frac{1}{x+4}\bigg)$$ one has $$\begin{eqnarray} \sum_{i=0}^{\infty}\frac{i!}{(i+4)!}&=&\sum_{i=0}^{\infty}\frac{1}{(i+1)(i+2)(i+3)(i+4)}\\ &=&-\frac12\sum_{i=0}^{\infty}\bigg(\frac{1}{i+2}-\frac{1}{i+3}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{i+1}-\frac{1}{i+4}\bigg)\\ &=&-\frac12\cdot\frac12+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{i+1}-\frac{1}{i+2}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{i+2}-\frac{1}{i+3}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{i+3}-\frac{1}{i+4}\bigg)\\ &=&-\frac14+\frac16(1+\frac12+\frac13)\\ &=&\frac1{18}. \end{eqnarray}$$
You can find this using telescoping sums. Observe he pattern: $$\frac{1}{i(i+1)} = \frac{1}{i}-\frac{1}{i+1}$$
$$\frac{1}{i(i+1)}-\frac{1}{(i+1)(i+2)}=\frac{2}{i(i+1)(i+2)}$$
$$\frac{1}{i(i+1)(i+2)}-\frac{1}{(i+1)(i+2)(i+3)}=\frac{3}{i(i+1)(i+2)(i+3)}$$
$$\frac{1}{i(i+1)(i+2)(i+3)}-\frac{1}{(i+1)(i+2)(i+3)(i+4)}=\frac{4}{i(i+1)(i+2)(i+3)(i+4)}$$
etc. In your case the ratio of factorials is just the product of four consecutive integers, so the third formula applies, giving the result $$\frac{1}{3} \frac{1}{1\cdot 2\cdot 3}$$