Question

How can this be done? Finding the radius of convergence and the interval of convergence of the series?

Find the radius of convergence and the interval of convergence of the series

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

Answer 1

Use the ratio test, to see that:

`lim_(n->oo) ((2^(n + 1)(x - 2)^(n + 1))/((n + 3)!))/((2^n(x -2)^n)/((n + 2)!)) < 1`

Since we know that if the ratio is between `0` and `1`, then the series converges (and we are looking for the values of `x` that ensure the series converges).

`2(x - 2) lim_(n-> oo) 1/(n + 3) < 1`

`2(x -2)(0) < 1`

Since this is true for all values of `x`, the radius of convergence is `oo` and the interval is `(-oo, oo)`.

Hopefully this helps!