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Radius of convergence of a power series

  • Thread starter jason177
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  • #1
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Homework Statement



Find the radius of convergence of

[tex]\sum[/tex]n!*xn from n=0 to [tex]\infty[/tex]

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 [tex]\infty[/tex] of abs(n+1). It seems to me that the radius of convergence would be 0 but the answer key says that it is [tex]\infty[/tex] so can someone help explain this.
 

Answers and Replies

  • #2
1,033
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the radius of convergence means what?

I cant 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.
 
  • #3
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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.
 
  • #4
lanedance
Homework Helper
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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
[tex] \sum a_n x^n [/tex]
the radius of convergence r, is the such that teh above powere series converges for all
[tex] |x|< r [/tex]

in you case
[tex] a_n = \frac{1}{n!}[/tex]

any way, back to your ratio test, i think you were pretty much there but missed a divide sign somehwere
[tex] \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}[/tex]

now simplify down
 
Last edited:
  • #5
lanedance
Homework Helper
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sorry - i read your post wrong, I agree with your original statement

[itex] \sum n!x^n [/itex] has a radius of convergence of zero
 
  • #6
1,033
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ohhhh right because it will only converge when x = 0. yup sorry for my confusion
 

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