Radius of convergence of a power series

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Homework Help Overview

The discussion revolves around finding the radius of convergence for the power series \(\sum n! x^n\) from \(n=0\) to \(\infty\). Participants are exploring the implications of the ratio test and the definition of radius of convergence in relation to this series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the ratio test and its results, with some questioning the interpretation of the radius of convergence. There are attempts to clarify the concept of radius of convergence and its relation to convergence of the series.

Discussion Status

There is an ongoing exploration of the radius of convergence, with some participants expressing confusion about the results of the ratio test and the implications for convergence. Guidance has been offered regarding the definition of radius of convergence, but no consensus has been reached on the correct interpretation.

Contextual Notes

Participants are grappling with the definitions and implications of convergence in the context of power series, and there are references to external resources for further reading. Some participants express uncertainty about their understanding of the concepts involved.

jason177
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Homework Statement



Find the radius of convergence of

\sumn!*xn from n=0 to \infty

Homework Equations




The Attempt at a Solution



I did the ratio test and was able to get it down to abs(x) * lim as n approaches \infty of abs(n+1). It seems to me that the radius of convergence would be 0 but the answer key says that it is \infty so can someone help explain this.
 
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the radius of convergence means what?

I can't quite remember, but something tells me that it is the distance from a to x, that the series converges at. If i can explain this properly, your x value there is (x-a) with an 'a' value of 0. So basically is is the distance from a that you will have to go until the series converges? But this series can never converge if you do your tests properly. That is why the distance is infinite.

If anyone would like to explain this better, please do as it has been a while and there are likely shaky parts of my explanation.
 
The radius of convergence is just the radius of x values in which the series will converge (basically just 1/2 the interval of convergence). So with a radius of convergence of infinity the series should converge for all values of x, or at least that's how I understood it when my teacher went over it.
 
you're pretty much right, for more reading have a look at http://en.wikipedia.org/wiki/Radius_of_convergence

when you have power series
\sum a_n x^n
the radius of convergence r, is the such that the above powere series converges for all
|x|< r

in you case
a_n = \frac{1}{n!}

any way, back to your ratio test, i think you were pretty much there but missed a divide sign somehwere
\frac{a_{n+1}}{a_n}= \frac{\frac{x^{n+1}}{(n+1)!}}{\frac{x^n}{n!}} = \frac{n! x^{n+1}}{(n+1)!x^n}

now simplify down
 
Last edited:
sorry - i read your post wrong, I agree with your original statement

\sum n!x^n has a radius of convergence of zero
 
ohhhh right because it will only converge when x = 0. yup sorry for my confusion
 

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