# convergence of a series

• Jan 6th 2010, 08:47 AM
Emma L
convergence 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 AM
Calculus26
You are correct
• Jan 6th 2010, 09:06 AM
Shanks
By the Stirling formula, $\frac{r!}{r^2e^r}$ does not tend to 0 as r approches infinity, thus the series diverges.