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Showing that a series diverges

by Sentral
Tags: diverges, series, showing
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Sentral
#1
Nov13-09, 07:15 PM
P: 6
1. The problem statement, all variables and given/known data

Show that the series [tex]\sum_{0}^{\infty}(2n)!/(n!)^2*(1/4)^n[/tex] diverges

2. Relevant equations

I don't know which convergence test to use

3. The attempt at a solution

I don't have one, because I don't know which convergence test to use. If someone can tell me what to use, I will be able to figure out this problem.
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LCKurtz
#2
Nov13-09, 07:58 PM
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Why don't you just try the most likely one?
Sentral
#3
Nov13-09, 08:01 PM
P: 6
I know that the basic ones such as ratio, divergence, alternating series, comparison, and integral don't work with this series. I believe it uses a test that I haven't learned about, so I was wondering what that could be.

LCKurtz
#4
Nov13-09, 08:12 PM
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Showing that a series diverges

I am just leaving for the evening so I don't have time to work on it myself. But I would be very surprised if the ratio test won't settle it. Did you try the root test?

I will check back later.
Mark44
#5
Nov13-09, 08:42 PM
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Quote Quote by Sentral View Post
I know that the basic ones such as ratio, divergence, alternating series, comparison, and integral don't work with this series. I believe it uses a test that I haven't learned about, so I was wondering what that could be.
No, one of the tests you mentioned will work. It's not an alternating series, so you can ignore that test. You wouldn't want to apply the integral test on this series, I don't think, so that eliminates that test.

Show us what you've done...
LCKurtz
#6
Nov14-09, 12:54 PM
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Clarification: Is that [itex](1/4)^n[/itex] in the numerator or the denominator of the fraction? I'm guessing the numerator, making the [itex]4^n[/itex] in the denominator, which makes the problem tougher than I thought at first glance. Is that right? I'm getting the ratio test fails too...
Sentral
#7
Nov14-09, 12:59 PM
P: 6
It's just the first fraction times (1/4)n. I guess there should be two parenthesis around the first fraction so it's just that quantity multiplied by the (1/4)n
jgens
#8
Nov14-09, 02:37 PM
P: 1,622
My initial response was deleted by the moderators, but you can prove that the sum diverges by considering the sequence [itex]a_n = 1/n[/itex].
Dick
#9
Nov14-09, 03:12 PM
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You can also learn things about series like this using Stirling's approximation.


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