Radius of convergence of a power series

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.

 

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

 

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

 

ohhhh right because it will only converge when x = 0. yup sorry for my confusion