## Calculus (3rd Edition)

The series $\Sigma_{n=1}^{\infty}4^na_n$ diverges.
Let $b_n=4^na_n$; then applying the ratio test, we have $$\rho=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_{n}}\right|=\lim _{n \rightarrow \infty} \frac{4^{n+1}}{4^n}\left|\frac{a_{n+1}}{a_{n}}\right|=\frac{4}{3}\gt1$$ Hence, the series $\Sigma_{n=1}^{\infty}4^na_n$ diverges.