## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty} \frac{2 n+1}{4^{n}}$$ Since $$2 n+1\lt2^n,\ \ \ \ \ \ \ n \geq 3$$ Then $$\frac{2 n+1}{4^{n}}\lt\frac{2^{n}}{4^{n}}=\left(\frac{1}{2}\right)^{n}$$ Since $\sum \left(\frac{1}{2}\right)^{n}$ is a convergent geometric series, then $\sum_{n=1}^{\infty} \frac{2 n+1}{4^{n}}$ also converges.