Find radius of convergence and interval of convergence for the series

Click For Summary
SUMMARY

The series in question is represented by the expression x^n/(2n-1), which has a radius of convergence of 1. The ratio test confirms that the absolute value of x must be less than or equal to 1 for convergence. While both endpoints, -1 and 1, yield convergence when plugged into the original series, the interval of convergence is [-1, 1) due to the divergence of the series at x=1, specifically when evaluated as ##\sum \frac{1}{2n-1}##. This divergence at x=1 necessitates the use of a soft bracket in the interval notation.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the ratio test for series
  • Knowledge of interval notation in mathematics
  • Basic calculus concepts, particularly limits and infinite series
NEXT STEPS
  • Study the comparison test for series convergence
  • Learn about different convergence tests, such as the root test
  • Explore the concept of power series and their convergence properties
  • Investigate the implications of endpoint behavior in series convergence
USEFUL FOR

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

emk
Messages
3
Reaction score
0
x^n/(2n-1) is the series. It starts at 1 and goes to infinity.

I did the ratio test on it and got abs.(x)

So the radius of convergence=1, and then I plugged -1 and 1 into the original series and got that they both converged. But the answer is [-1,1). Why aren't they both hard brackets?
 
Physics news on Phys.org
emk said:
x^n/(2n-1) is the series. It starts at 1 and goes to infinity.

I did the ratio test on it and got abs.(x)

So the radius of convergence=1, and then I plugged -1 and 1 into the original series and got that they both converged. But the answer is [-1,1). Why aren't they both hard brackets?

Pay special attention to x=1. When you plug it into your series you get ##\sum \frac{1}{2n-1}##.

You know that diverges...

Hint : Compare it to something you know diverges already.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K