Find all 
such that the series
![\[ \sum_{n=0}^{\infty} \frac{x^n}{(n+3)!} \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-614b221db427e6256296bd99e8d94d55_l3.png "Rendered by QuickLaTeX.com")
converges and compute the sum.
First, we apply the ratio test
![\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{|x|^{n+1}}{(n+4)!} \right) \left( \frac{(n+3)!}{|x|^n} \right)\\[9pt] &= \lim_{n \to \infty} \frac{|x|}{n+4} \\[9pt] &= 0 \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3031059f579d03455c852ae038986b1e_l3.png "Rendered by QuickLaTeX.com")
for all 
. Therefore this converges for all . Furthermore, we compute the sum, for 
(since we’ll want to divide by in here),
![\begin{align*} \sum_{n=0}^{\infty} \frac{x^n}{(n+3)!} &= \sum_{n=3}^{\infty} \frac{x^{n-3}}{n!} \\[9pt] &= \frac{1}{x^3} \sum_{n=3}^{\infty} \frac{x^n}{n!} \\[9pt] &= \frac{1}{x^3} \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} - 1 - x - \frac{x^2}{2} \right) \\[9pt] &= \frac{1}{x^3} \left( e^x - 1 - x - \frac{x^2}{2} \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8b4bf1209903438f5d4d12941af4005b_l3.png "Rendered by QuickLaTeX.com")
If 
then the sum is 0.