Trying to find the radius of convergence of this complicated infinite series

skyturnred
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Homework Statement



k is a positive integer.

\sum^{\infty}_{n=0} \frac{(n!)^{k+2}*x^{n}}{((k+2)n)!}

Homework Equations





The Attempt at a Solution



I have no idea.. this is too confusing. I tried the ratio test (which is the only way I know how to deal with factorials) but I get stuck at the following

lim n->\infty of | x(n+1)^{k+2}\frac{[(k+2)n]!}{[(k+2)(n+1)]!} |

I can't seem to find a way to cancel out the factorials in the fractional portion of that
 
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Here's a thought. k is just some positive integer, right? The issue of convergence of a series arises in the "tail" of the series - that is, if you want to look at the convergence of \sum_{i=0}^{\infty}\ a_n x^n, you could just as easily look at the convergence of \sum_{i=N}^{\infty}\ a_n x^n. With that in mind, try splitting the series into n < k and n > k to simplify things slightly. Also you can drop the absolute values around those factorials, since they're positive.
 
Try and focus on the problem by doing the special case k=0. Then try k=1. Can you generalize?
 
skyturnred said:
I can't seem to find a way to cancel out the factorials in the fractional portion of that

When I cancel factorials, I usually expand the factorial into maybe three or more factors, e.g. n!=n(n-1)(n-2)... and (n-1)!=(n-1)(n-2)...

Then it is clear that \frac{n!}{(n-1)!}=n
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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