We went through an example in class the other week but I don't quite follow.
Basically, we are given a power series from k=0 to infinity: [(k!)^3 / (3k!)!]*x^k
Using the ratio test: mod of bk+1/bk eventually leads to:
[(k+1)^3 * mod x] / [(3k+3)(3k+2)(3k+1)]
Then this is meant to converge to [mod x] / 27, thus giving a radius of convergence of 27.
But I don't see how you get that it converges to [mod x]/27.
Any help please?
power series - radius of convergence
- 01-04-2012 12:46
- 01-04-2012 12:50
(3k+1), (3k+2), (3k+3) ~3k each as k--> infinity
k+1~k as k-->infinity
So you get approximately k^3|x|/(3k)^3
(I don't know how much precision you want here, so if you want slightly more rigour, I'm sure that's possible)
- 01-04-2012 13:04
Yeah I guess I can see how that one works. But then for the next example (which the lecturer didn't go through so I did this working out on my own, I don't think I quite get it.
I have a power series again from k=0 to infinity. [(-1)^k / 2k!] * x^2k
Using the same method again, it brings me to x^2 /(2k+1)(2k+2).
This example is supposed to have a radius of convergence of 30. So maybe I did something wrong in my calculation?
- 01-04-2012 13:55
I agree with the x^2/(2k+1)(2k+2) bit. I'm not sure how they're getting to 30 from there either.
- 10 followers
- 0 badges
- 01-04-2012 14:17
(Original post by james.h)
- 01-04-2012 14:37
It's a Maclaurin series, isn't it (exponential)? I've no idea where the value 30 comes from for the radius of convergence, since it ought to converge for all real x.
- 01-04-2012 14:55