Does the series SIGMA (0 to infinity) r!/r^2(e^r) converge or diverge and why?

Tried using the ratio test and got to lim (r tend to inf) r^2/(r+1)e , which equals infinity(?). sooo it diverges as its greater than 1?

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- Jan 6th 2010, 08:47 AMEmma Lconvergence of a series
Does the series SIGMA (0 to infinity) r!/r^2(e^r) converge or diverge and why?

Tried using the ratio test and got to lim (r tend to inf) r^2/(r+1)e , which equals infinity(?). sooo it diverges as its greater than 1? - Jan 6th 2010, 08:51 AMCalculus26
You are correct

- Jan 6th 2010, 09:06 AMShanks
By the Stirling formula, $\displaystyle \frac{r!}{r^2e^r}$ does not tend to 0 as r approches infinity, thus the series diverges.