## Calculus (3rd Edition)

The series $\Sigma_{n=1}^{\infty}2^na_n$ converges.
Let $b_n=2^na_n$, then applying the ratio test, we have $$\rho=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_{n}}\right|=\lim _{n \rightarrow \infty} \frac{2^{n+1}}{2^n}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{2}{3}\lt 1$$ Then the series $\Sigma_{n=1}^{\infty}2^na_n$ converges.