<s>[INST] Radius and Interval of Convergence of $\sum (-1)^n \frac{ (x+2)^n }{n2^n}$ I have found the radius of convergence $R=2$ and the interval of convergence $I =[-2,2)$ for the following infinite series:
$\sum_{n=1}^\infty (-1)^n \frac{ (x+2)^n }{n2^n}$
Approach:
let
$a_n = (-1)^n \frac{ (x+2)^n }{n2^n}$
Take the ratio test
$\lim_{x\to -\infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| -\frac{1}{2}(x+2)\right|$
So it converges for
$-2 < (x+2) < 2, R = 2$
Test end points
if $x = -2$
$\sum_{n=1}^\infty (-1)^n(0) =0 < \infty$ , converges
if $x = 2$
$\sum_{n=1}^\infty (-1)^n \frac{4^n}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^{2n}}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^n}{n}$
let $b_n = \frac{2^n}{n}$
$b_{n+1} > b_n$ for every value of $n$
By the Alternating Series Test, $\sum_{n=1}^\infty (-1)^n b_n$ diverges
Hence,
$R = 2$ and
$I = [-2, 2)$
Is my approach correct? Are there any flaws, or better ways to solve?
 [/INST] The radius of convergence is right. The interval of convergence started out OK, you wrote that for sure we have convergence if $-2\lt x+2\lt 2$. But this should become $-4\lt x\lt 0$.
The endpoint testing was inevitably wrong, since the incorrect endpoints were being tested. 
We have convergence at $x=0$ (alternating series) and divergence at $x=-4$ (harmonic series).
</s>