(Check my work) Find the interval and radius of convergence of the following series?

I'll be adding my work in an answer below, so you don't have to respond immediately.

a. \sum_(n=1)^\infty((x+2)^n)/(n4^n)
b. \sum_(n=1)^\infty(2^n(x-2)^n)/((n+2)!)

1 Answer
Apr 24, 2018

See explanation (work is summarized)

Explanation:

For (a):

\color(red)\text(please check the brackets on my interval of convergence?)

a_n=(x+2)^n/(n4^n) and a_(n+1)=((x+2)^n(x+2))/((n+1)4^n(4))
\lim_(n\rarr\infty)|a_(n+1)/a_n|=...|x+2|/4
Simplifying from |x+2|/4\lt1, I got -6\ltx\lt2
The ROC is (2-(-6))/2=8/2=4.

But I am stuck on testing the endpoints.
x=-6\rArr\sum(-6+2)^n/(n4^n)=\sum(-4)^n/(n4^n) apply AST (true on both counts), convergent
x=2=rArr\sum(2+2)^n/(n4^n)=\sum4^n/(n4^n)=\sum1/n, harmonic series is divergent

IOC: [-6,2)

For (b):

\color(red)\text(please check over-all)

a_n=(2^n(x-2)^n)/(n+2)! and a_(n+1)=(2^n(2)(x-2)^n(x-2))/((n+3)(n+2)!
\lim_(n\rarr\infty)|a_(n+1)/a_n|=\lim_(n\rarr\infty)|(2(x-2))/(n+3)|=0\lt1

The sequence converges by Ratio Test. Thus, ROC is \infty and IOC is (-\infty,\infty)


(corrections from tutor for part B)

a_n = (2^n(x-2)^n)/((n+2)!)

a_(n+1) = (2^(n+1)(x-2)^(n+1))/(((n+1)+2)!) = (2 \ 2^n(x-2)(x-2)^n)/((n+1) \ (n+2)!)

a_(n+1) /a_n = (2 \ 2^n(x-2)(x-2)^n)/((n+1) \ (n+2)!) * ((n+2)!)/(2^n(x-2)^n) = (2 \ (x-2))/(n+1)