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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You are correct
By the Stirling formula, $\displaystyle \frac{r!}{r^2e^r}$ does not tend to 0 as r approches infinity, thus the series diverges.
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