Answer
Converges
Work Step by Step
Given$$\sum_{n=1}^{\infty} 4^{-2 n+1}=\sum_{n=1}^{\infty} 4\left(4^{-2}\right)^{n}$$
Then
$$\sum_{n=1}^{\infty} 4^{-2 n+1}=\sum_{n=1}^{\infty} 4 \frac{1}{16^{n}}$$
which is a geometric series with $c=4$ and $r=\frac{1}{16}<1 $. Thus, the given series converges.
