## Calculus (3rd Edition)

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.