## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}\frac{n^{5}}{2^{n}}$$ Since for $n\geq1$ \begin{aligned} n^{5} & \leq\left(\frac{3}{2}\right)^{n} \\ \frac{n^{5}}{2^{n}} & \leq\left(\frac{3}{2}\right)^{n} \\ \frac{n^{5}}{2^{n}} & \leq\left(\frac{3}{2 \times 2}\right)^{n} \\ \frac{n^{5}}{2^{n}} & \leq\left(\frac{3}{4}\right)^{n} \end{aligned} Compare with $\displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{4}\right)^{n}$, a convergent geometric series $|r|<1$; then $\displaystyle\sum_{n=1}^{\infty}\frac{n^{5}}{2^{n}}$ also converges.