Answer
The ratio test is inconclusive and the series $\Sigma_{n=1}^{\infty}2^na_n$ maybe diverge or converge.
Work Step by Step
Let $b_n=3^na_n$; then applying the ratio test, we have
$$
\rho=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_{n}}\right|=\lim _{n \rightarrow \infty} \frac{3^{n+1}}{3^n}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{3}{3}= 1
$$
In this case, the ratio test is inconclusive and the series $\Sigma_{n=1}^{\infty}2^na_n$ maybe diverge or converge.
