Tricky series radius of convergence question (analysis course)

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Homework Help Overview

The discussion revolves around finding the radius of convergence for the series defined by the sum from 1 to n of 1/(n^n) * x^(2^n). The subject area pertains to series convergence within an analysis course.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the applicability of different convergence tests, such as the ratio test and the nth root test. There is uncertainty regarding how to handle the 2^n exponent in the series. Questions arise about the conditions under which the limsup of the expression is less than one, and participants express doubts about the validity of certain values of |x|.

Discussion Status

The discussion is ongoing, with participants sharing various approaches and considerations. Some guidance is offered regarding the use of l'Hôpital's rule and comparison tests, but no consensus has been reached on the analytical proof of the radius of convergence.

Contextual Notes

Participants note challenges in proving the behavior of the series at specific values of |x|, particularly at |x|=1, and discuss the implications of these cases on convergence.

pcvt
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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
 
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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?
 
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