What Is the Radius of Convergence for This Series?

Click For Summary
SUMMARY

The radius of convergence (ROC) for the series \(\sum_{n=0}^{\infty}\frac{(n!)^3}{(3n)!}z^{3n}\) can be determined using the ratio test. The series can be transformed into \(f(z) = \sum_{n=0}^{\infty}\frac{(n!)^3}{(3n)!}z^{n}\), which converges for \(|z| < R\) and diverges for \(|z| > R\). Consequently, the original series converges when \(|z|^3 < R\), leading to the conclusion that the ROC is defined by the inequality \(|z| < R^{1/3}\).

PREREQUISITES
  • Understanding of power series and their convergence properties
  • Familiarity with the ratio test for series convergence
  • Knowledge of factorial functions and their growth rates
  • Basic concepts of limits, specifically limsup
NEXT STEPS
  • Study the application of the ratio test in detail, particularly for series with factorial terms
  • Explore the concept of limsup and its role in determining convergence
  • Investigate other methods for finding the radius of convergence, such as the root test
  • Examine the implications of variable transformations in power series
USEFUL FOR

Students and educators in mathematics, particularly those focusing on complex analysis and series convergence, as well as anyone seeking to deepen their understanding of power series and convergence tests.

jessicaw
Messages
54
Reaction score
0

Homework Statement


find the roc of:
[tex]\sum_{n=0}^{\infty}\frac{(n!)^3}{(3n)!}z^{3n}[/tex]


Homework Equations


limsup
ratio test


The Attempt at a Solution


i think use of limsup is quite difficult as factorial there.
but i do not know how to use the ratio test because z is another variable. It means there are two independent variable, but ratio test is for one variable(i.e. n)only, rite?


Also besides ratio test and limsup, are there other methods to find the "roc"?
thx!
 
Physics news on Phys.org
No the ratio test certainly still works. The intuitive way to approach it is to recall what the radius of convergence actually means. If you had just z^n instead of z^(3n), then applying the ratio test you get the radius convergence R so that the series
[tex] f(z) = \sum_{n=0}^{\infty}\frac{(n!)^3}{(3n)!}z^{n}[/tex]
converges for |z| < R, diverges for |z| > R. But the series you have is simply f(z^3), so you have convergence when |z|^3 < R, and divergence of |z|^3 > R. So what is the radius of convergence for the power series with the z^(3n) term?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
2K