# 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)!)

Jun 4, 2018

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).

$2 \left(x - 2\right) {\lim}_{n \to \infty} \frac{1}{n + 3} < 1$

$2 \left(x - 2\right) \left(0\right) < 1$

Since this is true for all values of $x$, the radius of convergence is $\infty$ and the interval is $\left(- \infty , \infty\right)$.

Hopefully this helps!