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# power series - radius of convergence Watch

1. 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.

2. (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
Make sense?
(I don't know how much precision you want here, so if you want slightly more rigour, I'm sure that's possible)
3. 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?
4. I agree with the x^2/(2k+1)(2k+2) bit. I'm not sure how they're getting to 30 from there either.

5. 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.
6. (Original post by james.h)

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.
No, it's
7. (Original post by Mathlete29)
No, it's
It's cos, so convergence is still as he said.

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