Radius of convergence of a power series

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SUMMARY

The radius of convergence for the power series \(\sum n! x^n\) is determined to be 0, indicating that the series converges only at \(x = 0\). The ratio test was applied, leading to the conclusion that the limit of the ratio of successive terms approaches infinity, which confirms the radius of convergence is indeed 0. This means that the series diverges for all other values of \(x\). The discussion clarifies that the radius of convergence represents the distance from the center point \(a\) (which is 0 in this case) within which the series converges.

PREREQUISITES
  • Understanding of power series and their general form
  • Familiarity with the ratio test for convergence
  • Knowledge of factorial notation and its properties
  • Basic concepts of limits in calculus
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  • Study the concept of radius of convergence in various power series
  • Explore examples of power series with different radii of convergence
  • Learn about the implications of convergence and divergence in series
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Students studying calculus, particularly those focusing on series and convergence, as well as educators looking for clear explanations of power series behavior.

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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