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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

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    $\begingroup$ Your ratio test is flipped! $\endgroup$ – DaveNine Aug 25 '17 at 2:31
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    $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Aug 25 '17 at 7:04
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You idea is correct but you computed the limit of the inverse of the quotient in ratio test.

This is the real relation:

$$\frac{U_{n+1}}{U_n} \rightarrow 0<1$$ thus the series converges.

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$$\begin{align}\sum U_n=\sum_{n=1} ^\infty \frac{n^3-n+1}{n!}\\=\sum_{n=1} ^\infty \frac{(n-1)n(n+1)}{n!}+\sum_{n=1} ^\infty\frac{1}{n!}\\=\sum_{n=2} ^\infty\frac{n+1}{(n-2)!}+\sum_{n=1} ^\infty\frac{1}{n!}\\=\sum_{n=3} ^\infty\frac{1}{(n-3)!}+\sum_{n=2} ^\infty\frac{3}{(n-2)!}+\sum_{n=1} ^\infty\frac{1}{n!}\end{align}$$

Clearly you can see each of these sums converges.

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