Question

What is the interval of convergence of #sum_1^oo n!(2x-1)^n #?

Answer 1

Converges only for `x = 1/2`

Explanation:

`sum_1^oo n!(2x-1)^n equiv sum_1^oo n!y^n` with `y = 2x-1`

With `y = y_0 ne 0` we have

`lim_{n->oo}n!abs(y_0)^n= oo` because `EE n_0 < n and n_0 > abs(y_0)`

Concluding, the convergence radius comprehends only the point `y = 0` or `2x-1=0->x = 1/2`