Tricky series radius of convergence question (analysis course)

Click For Summary
SUMMARY

The radius of convergence for the series defined by the sum from 1 to n of 1/(n^n) * x^(2^n) is determined to be |x| <= 1. The nth root test is applicable in this scenario, leading to the conclusion that for |x| < 1, the series converges, while for |x| > 1, it diverges. The analytical proof involves demonstrating that the limit superior of 1/n * x^(2^n/n) is less than one, particularly using techniques such as l'Hôpital's rule and comparison tests.

PREREQUISITES
  • Understanding of series convergence tests, specifically the nth root test and ratio test.
  • Familiarity with limit superior concepts in real analysis.
  • Knowledge of exponential functions and their growth rates.
  • Basic proficiency in calculus, including l'Hôpital's rule.
NEXT STEPS
  • Study the application of the nth root test in series convergence.
  • Research limit superior and its implications in real analysis.
  • Learn about l'Hôpital's rule and its use in evaluating limits.
  • Explore comparison tests for series convergence, particularly in relation to exponential growth.
USEFUL FOR

Students in analysis courses, mathematicians focusing on series convergence, and educators teaching advanced calculus concepts.

pcvt
Messages
17
Reaction score
0

Homework Statement


Find the radius of convergence of sum from 1 to n of

1/(n^n) * x^(2^n)


Homework Equations





The Attempt at a Solution


Clearly ratio test isn't going to work straight away. I'm not sure how to deal with the 2^n exponent
 
Physics news on Phys.org
What about the nth root test?
 
From there I get down to finding for what values the limsup of 1/n x^(2^n/n) is less than one for

I believe its |x|<=1 but I'm not sure how to analytically prove this
 
pcvt said:
From there I get down to finding for what values the limsup of 1/n x^(2^n/n) is less than one for

I believe its |x|<=1 but I'm not sure how to analytically prove this

I don't think |x|=1 works. But the case |x|<1 is easy. And |x|>1 is not much harder, you just want to show it diverges somehow. I'd say the easiest way is to combine l'Hopital or a ratio test with a comparison test. For example, can you show 2^n/n>n for large enough values of n?
 
Last edited:

Similar threads

Interval of convergence of power series from ratio test
  • · Replies 2 ·
Replies
2
Views
2K
Does this series converge uniformly?
  • · Replies 7 ·
Replies
7
Views
2K
Proving convergence and divergence of series
  • · Replies 3 ·
Replies
3
Views
2K
Addition of power series and radius of convergence
  • · Replies 14 ·
Replies
14
Views
2K
What is the radius of convergence for a series with logarithmic terms?
  • · Replies 3 ·
Replies
3
Views
2K
On the definition of radius of convergence; a small supremum technicality
  • · Replies 7 ·
Replies
7
Views
2K
Convergence of a Series: Radius and Endpoints
  • · Replies 10 ·
Replies
10
Views
2K
Valid conclusion for an absolutely convergent sequence
  • · Replies 4 ·
Replies
4
Views
1K
On the ratio test for power series
  • · Replies 2 ·
Replies
2
Views
2K
Set of convergence of a Power series
  • · Replies 4 ·
Replies
4
Views
2K