# How do I test the Convergence/ divergence of this series?

Given series is $$\sum U_n=\sum_{n=1} ^\infty \frac{n^3-n+1}{n!}$$ I need to test its convergence.

I thought about using ratio test for the same but I am stuck on how to proceed after I reach a step where:

$$\lim_{n\to\infty}\frac{U_n}{U_{n+1}}=\lim_{n\to\infty}\frac{(n^3-n+1)(n+1)}{(n+1)^3-n}=\to\infty$$

How do I proceed from here? Some guidance would be appreciated

• Your ratio test is flipped! – DaveNine Aug 25 '17 at 2:31
• This is a overly trivial question for a ~4k user. You should be able to check by yourself that $\sum_{n\geq 1}\frac{n^k}{n!}$ is an absolutely convergent series for any $k\in\mathbb{N}$. Maybe compute its value, too. – Jack D'Aurizio Aug 25 '17 at 7:04

$$\frac{U_{n+1}}{U_n} \rightarrow 0<1$$ thus the series converges.
\begin{align}\sum U_n=\sum_{n=1} ^\infty \frac{n^3-n+1}{n!}\\=\sum_{n=1} ^\infty \frac{(n-1)n(n+1)}{n!}+\sum_{n=1} ^\infty\frac{1}{n!}\\=\sum_{n=2} ^\infty\frac{n+1}{(n-2)!}+\sum_{n=1} ^\infty\frac{1}{n!}\\=\sum_{n=3} ^\infty\frac{1}{(n-3)!}+\sum_{n=2} ^\infty\frac{3}{(n-2)!}+\sum_{n=1} ^\infty\frac{1}{n!}\end{align}