Question

(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)!)`

Answer 1

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)`

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(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)`