Finding if a series is convergent.

In summary, the conversation covers a problem of finding the values of K for which a given series is convergent. The method used is the ratio test, and it is determined that the series diverges for any possible K. However, the conversation also suggests considering Stirling's Approximation or using mathematical induction to cover all integer values of K greater than 2.
  • #1
kelumhi
2
0
Finding if a series is convergent-Answered

Homework Statement


Find for which values of K is the fallowing series convergent.

[tex]\sum[/tex]((n!)2)/((kn)!)

where:
N is the variable.
K is a constant or a list of constant (eg. "(2,91]")

Homework Equations



I believe the ratio test, which states that if (f(n+1)/(f(n) as n approaches infinity is less then 1, it converges.

The Attempt at a Solution



I believe the obvious way to go about this would be the ratio test which is as fallows"

[PLAIN]http://img688.imageshack.us/img688/2140/equation1.png
[PLAIN]http://img146.imageshack.us/img146/2783/equation2.png
[PLAIN]http://img232.imageshack.us/img232/1554/equation3.png

1>((n+1)!*(n+1)!/(k(n+1))!*((Kn)!/(n!*n!) as n[tex]\rightarrow[/tex] [tex]\infty[/tex]

1>(n+1)(n+1)/(k(n+1)) as n[tex]\rightarrow[/tex] [tex]\infty[/tex]

1> (n+1)/k

which is not true, therefore this series must diverge for any possible K.

my question: am i doing anything wrong or did the teacher give a trick question?
 
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  • #2
actually sorry, I take that back, i think there is a mistake
[tex] \frac{ (kn)! }{ (k(n+1))! } = \frac{ (kn)! }{ (kn+k)! } \neq \frac{1}{k(n+1)} [/tex]
 
Last edited:
  • #3
as a guide, try k = 1,2,3 and see whether you get a converging series

another way do this covering all integer k>1 would be to use Stirlings's Approximation for n! as n gets large
 
  • #4
Oh, Thanks for your help lanedance. i found it pretty intuitive that 2+ was the answer but i wouldn't have thought to be able to solve it mathematically i would need to distribute the K then simply insert a value. you've been very helpful.
 
  • #5
to be rigorous, you need to show its true for all k>2, not just one larger than 2 (3)

so options are
- have a look at Stirlings Approximation [itex] n! \approx n^{n}e{-n}\sqrt{2\pi n }[/itex]
- try mathematical induction

i haven't tried either fully but think both should work
 

FAQ: Finding if a series is convergent.

1. How do you determine if a series is convergent?

To determine if a series is convergent, you can use various methods such as the comparison test, the ratio test, or the root test. These tests involve comparing the given series to a known convergent or divergent series and analyzing the behavior of the terms in the series.

2. What is the comparison test and how does it work?

The comparison test is a method of determining if a series is convergent by comparing it to a known convergent or divergent series. If the given series has terms that are always smaller than the terms in the known convergent series, then the given series is also convergent. If the given series has terms that are always larger than the terms in the known divergent series, then the given series is also divergent.

3. How do the ratio and root tests work?

The ratio and root tests are methods of determining the convergence of a series by analyzing the behavior of the terms in the series. The ratio test involves taking the limit of the ratio of consecutive terms in the series. If the limit is less than 1, then the series is convergent. The root test involves taking the limit of the nth root of the absolute value of the terms in the series. If the limit is less than 1, then the series is convergent.

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

A series is absolutely convergent if the series of the absolute values of its terms is convergent. A series is conditionally convergent if the series of its terms is convergent, but the series of the absolute values of its terms is divergent. In other words, the signs of the terms in a conditionally convergent series alternate, leading to a cancellation effect that makes the series convergent.

5. Can a series be both absolutely and conditionally convergent?

No, a series can only be either absolutely or conditionally convergent. If a series is absolutely convergent, it is also conditionally convergent. However, if a series is conditionally convergent, it may not be absolutely convergent. This is because the cancellation effect in a conditionally convergent series may be lost when taking the absolute values of its terms.

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