Using Ratio Test to Decide Convergent/Divergent

McAfee · Dec 8, 2011

Homework Statement

Online homework problem: series sum from n=1 to infinity of a_n, with a_n = e^(n+3) divided by the square root of (n+5(n+2)!), asking the reader to compute the Ratio Test limit L and choose a true statement from options A to F.

The Attempt at a Solution

Handwritten ratio test working on lined paper: a_n = e^(n+3) / (sqrt(n+5)(n+2)!), then the limit as n goes to infinity of the ratio a_(n+1)/a_n, simplified to e sqrt(n+5)(n+1)(n+2) / (sqrt(n+6)(n+3)!) with cancellations marked.

I got that the limit goes to infinity using the Ratio Test and that would mean B. But I'm not 100% percent sure. Just looking for some to verify my answer if I'm correct or not.

SammyS · Dec 8, 2011

McAfee said:

Homework Statement

The Attempt at a Solution

I got that the limit goes to infinity using the Ratio Test and that would mean B. But I'm not 100% percent sure. Just looking for some to verify my answer if I'm correct or not.

Your limit is wrong, because your ratio is wrong.

[itex]\displaystyle\frac{(n+2)!}{(n+3)!}=\frac{1}{n+3}[/itex]

McAfee · Dec 8, 2011

SammyS said:

Your limit is wrong, because your ratio is wrong.

[itex]\displaystyle\frac{(n+2)!}{(n+3)!}=\frac{1}{n+3}[/itex]

Wow. I haven't used factorials in a while and forgot how they were used. Now looking at the problem I get L=0. That tells me that the series converges absolutely.

Thanks a lot. There are so many things is math to remember when solving a problem.

Using Ratio Test to Decide Convergent/Divergent

Homework Statement

The Attempt at a Solution

Discussion

Homework Statement

The Attempt at a Solution

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