Answer
The series converges for all $x$ in the interval $[-1/ 2,1/2)$.
Work Step by Step
We apply the ratio test
$$ \rho=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim _{n \rightarrow \infty} \frac{2^{(n+1)}x^{n+1}/(n+1) }{2^nx^n/n}=2|x|\lim _{n \rightarrow \infty} \frac{n}{n+1}=2|x| $$
Hence, the series $\Sigma_{n=0}^{\infty} 2^nx^n/n$ converges if and only if $|x|\lt1/2$. That is, the interval of convergence is $(-1/2,1/2)$.
Now, we check the end points:
At $x=-1/ 2$, then
$\Sigma_{n=0}^{\infty} 2^nx^n/n=\Sigma_{n=0}^{\infty} (-1)^n/n$ , converges (Alternating Series Test)
At $x=1/ 2$, then
$\Sigma_{n=0}^{\infty} 2^nx^n/n=\Sigma_{n=0}^{\infty}1/n$, diverges (P-Series Test)
Hence, the series converges for all $x$ in the interval $[-1/ 2,1/2)$.
