What is the radius of convergence of the power series (((2n)!x^(n))/((2n-1)!)), and what is its interval of convergence?

I used the ratio test and found that the radius of convergence is 0, as it is impossible for the absolute value of infinity to be less than 1. I am not sure I found the radius correctly though, as I am asked what the interval of convergence is. This confuses me because it doesn't seem that the power series converges at all.

Thanks!

To find the radius of convergence and interval of convergence of a power series, you used the ratio test, which is a good approach. The ratio test tells you whether a power series converges absolutely, which means it converges regardless of the sign of the terms.

However, in this case, your expression for the power series is incorrect. The power series you provided is:

∑ ((2n)!x^(n))/((2n-1)!)

The ratio test for convergence of a power series is given by the formula:

lim | (a_(n+1))/(a_n) | as n approaches infinity

To apply the ratio test to your series, we need to compute the limit:

lim n->∞ | ((2(n+1))!x^(n+1))/((2(n+1)-1)!)) | / | ((2n)!x^n)/((2n-1)!) |

Simplifying this expression, we get:

lim n->∞ | ((2n+2)!x^(n+1))/((2n+1)!)) | / | ((2n)!x^n)/((2n-1)!) |

Now, let's cancel out common terms and simplify further:

lim n->∞ | (2n+2)(2n+1)x^(n+1) / (2n)(2n-1)x^n |

After further simplification, we see that many terms cancel out:

lim n->∞ | (2n+2)(2n+1)x / (2n)(2n-1) |

Now, we can rewrite this expression as:

lim n->∞ | (2n+2)(2n+1) / (2n)(2n-1) | * | x |

Taking the limit as n approaches infinity, we see that the ratio simplifies to:

| 4x |

Now, we need to set this ratio less than 1 to determine the radius of convergence:

| 4x | < 1

Simplifying further, we get:

-1/4 < x < 1/4

This inequality gives us the interval of convergence. Therefore, the radius of convergence is 1/4, and the interval of convergence is (-1/4, 1/4).