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?

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).
