Power Series Converge Absolutely

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Homework Help Overview

The discussion revolves around determining the values of x for which the power series \(\sum^{∞}_{n=1} \frac{4^n * x^n}{n!}\) converges absolutely. Participants are exploring the application of the Ratio Test in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the Ratio Test and the implications of the limit being zero. Questions are raised about the significance of this limit in relation to the radius of convergence.

Discussion Status

The conversation includes attempts to clarify the results of the Ratio Test and how they relate to the convergence of the series. Some participants suggest that the series converges for all values of x based on the limit, while others question the need for further proof or validation of this conclusion.

Contextual Notes

There is mention of previous discussions regarding similar limits, indicating a potential confusion or need for deeper understanding of the concepts involved in convergence and the Ratio Test.

whatlifeforme
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Homework Statement


for what values of x does the series converge absolutely?

Homework Equations


[itex]\displaystyle \sum^{∞}_{n=1} \frac{4^n * x^n}{n!}[/itex]

The Attempt at a Solution


Ratio Test

[itex]\displaystyle \frac{4^{n+1} * x^{n+1}}{n+1)!} * \frac{n!}{4^n * x^n}[/itex]
4x * limit (n->inf) [itex]\displaystyle \frac{1}{n+1} = 0[/itex]

What do I do now since the limit is zero? I asked a similar question in another thread, but the limit turned out to be 1. I am sure that this limit is zero.
 
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What was your purpose in calculating that? How is that number, whether it is 0 or 1 or whatever, tell you about the "radius of convergence"?
 
whatlifeforme said:

Homework Statement


for what values of x does the series converge absolutely?


Homework Equations


[itex]\displaystyle \sum^{∞}_{n=1} \frac{4^n * x^n}{n!}[/itex]


The Attempt at a Solution


Ratio Test

[itex]\displaystyle \frac{4^{n+1} * x^{n+1}}{n+1)!} * \frac{n!}{4^n * x^n}[/itex]
4x * limit (n->inf) [itex]\displaystyle \frac{1}{n+1} = 0[/itex]

What do I do now since the limit is zero? I asked a similar question in another thread, but the limit turned out to be 1. I am sure that this limit is zero.

It means that the series converges for all values of x.

BTW, when you use the Ratio Test, you should be working with the absolute values of the terms in your series.

This is the ratio you should be working with:
$$ \frac{4^{n+1} * |x|^{n+1}}{(n+1)!} * \frac{n!}{4^n * |x|^n}$$
The result you get will be 4|x| ## \lim_{n \to \infty} 1/(n + 1)## = 0, which places no limits on the values of x.
 
do i leave the answer at 4|x| limn→∞1/(n+1) = 0 and put -inf < x < inf. or do i need to prove this somehow?
 

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