Showing that a series diverges

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In summary, the homework statement is that the series \sum_{0}^{\infty}(2n)!/(n!)^2*(1/4)^n diverges. The attempt at a solution is to try the most likely convergence test, which is the ratio test. However, the problem becomes tougher than it seems at first glance because the sum has a fraction in it. Another way to solve this problem is using the Stirling's approximation.
  • #1
Sentral
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Homework Statement



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

Homework Equations



I don't know which convergence test to use

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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  • #2
Why don't you just try the most likely one?
 
  • #3
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.
 
  • #4
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.
 
  • #5
Sentral said:
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...
 
  • #6
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...
 
  • #7
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
 
  • #8
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].
 
Last edited:
  • #9
You can also learn things about series like this using Stirling's approximation.
 

1. How do you show that a series diverges?

To show that a series diverges, you can use one of several tests such as the Divergence Test, the Integral Test, or the Comparison Test. These tests involve examining the behavior of the terms in the series and determining if they approach infinity or if they do not converge to a finite limit.

2. Can a series diverge if its terms approach zero?

Yes, a series can still diverge even if its terms approach zero. This is because the terms may decrease at a slower rate than the rate at which they approach zero, resulting in an infinite sum. An example of this is the Harmonic Series, which has terms that approach zero but the series still diverges.

3. How does the Divergence Test work?

The Divergence Test is a basic test for divergence that states that if the limit of the terms in a series does not approach zero, then the series must diverge. This is because if the terms do not approach zero, then they cannot converge to a finite sum. However, if the limit of the terms is zero, the test is inconclusive and other tests must be used.

4. What is the difference between absolute and conditional convergence?

A series is absolutely convergent if the sum of the absolute values of its terms is finite. A series is conditionally convergent if it is convergent, but not absolutely convergent. This means that the series converges, but if the signs of the terms were changed, the series would diverge. An example of a conditionally convergent series is the Alternating Harmonic Series.

5. Can a series diverge if its terms alternate between positive and negative values?

Yes, a series can still diverge even if its terms alternate between positive and negative values. This is because the terms may not approach zero quickly enough, resulting in an infinite sum. An example of this is the Alternating Harmonic Series, where the terms alternate between positive and negative but the series still diverges.

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