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

In summary, the given equation involves a series with a positive integer k that can be simplified by splitting it into two parts and using the factorial expansion method to cancel out the factorials. The ratio test can also be used to determine convergence.
  • #1
skyturnred
118
0

Homework Statement



k is a positive integer.

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

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

I can't seem to find a way to cancel out the factorials in the fractional portion of that
 
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  • #2
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 [itex]\sum_{i=0}^{\infty}\ a_n x^n[/itex], you could just as easily look at the convergence of [itex]\sum_{i=N}^{\infty}\ a_n x^n[/itex]. 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.
 
  • #3
Try and focus on the problem by doing the special case k=0. Then try k=1. Can you generalize?
 
  • #4
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. [tex]n!=n(n-1)(n-2)...[/tex] and [itex](n-1)!=(n-1)(n-2)...[/itex]

Then it is clear that [itex]\frac{n!}{(n-1)!}=n[/itex]
 

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

1. What is a radius of convergence?

A radius of convergence is a value that indicates the range of values for which a given infinite series will converge. It is the distance from the center of the series to the point where the series converges.

2. How is the radius of convergence determined?

The radius of convergence is determined by using the ratio test, which involves taking the limit of the absolute value of the ratio of consecutive terms in the series. If this limit is less than 1, then the series will converge within a certain radius. If the limit is greater than 1, then the series will diverge. If the limit is exactly 1, further tests are needed to determine the convergence or divergence of the series.

3. Can a series have more than one radius of convergence?

Yes, it is possible for a series to have multiple radii of convergence. This can occur when the series has different behaviors for different values of the independent variable. For example, a power series could have different radii of convergence for different values of x.

4. What happens if the radius of convergence is 0?

If the radius of convergence is 0, it means that the series will only converge at the center point. This could indicate that the series has a singularity at the center point or that the series itself is not defined at that point.

5. Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. The radius is a distance and therefore must be a positive value. If the ratio test yields a negative value, it is likely a sign error and the absolute value should be taken to determine the radius.

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