Question

What is the interval of convergence of `sum_1^oo [(2n)!x^n] / (n!)^2`?

Answer 1

`abs(x) < 1/4`

Explanation:

`sum_(k=0)^oo [(2k)!x^k] / (k!)^2=sum_(k=0)^oo ((2k),(k))x^k equiv 1/sqrt(1-4x)` for `absx < 1/4`

So the interval of convergence is given by `1-4x > 0` or for the series representation, `abs(x) < 1/4`