Question

Sum from n=1 to infinity that [(n!x^n)/[1*3*5...(2*n-1)]]. Find the radius of convergence and interval of convergence?

Answer 1

We have:

`sum_(n = 1)^oo (n!x^n)/((2n - 1)!)`

The radius of convergence is given by the ratio test.

`L = lim_(n->oo) (((n + 1)!x^(n + 1))/((2(n + 1) - 1)!))/((n!x^n)/((2n - 1)!)`

`L = |x|lim_(n-> oo) (n + 1)/((2n + 1)(2n))`

`L = |x| 0`

This means that the series converges for all values of `x`, or the interval of convergence is `(-oo, oo)`.

Hopefully this helps!