# 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)!)#

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

##### 1 Answer

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

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

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

Since this is true for all values of

Hopefully this helps!