Answer
Converges
Work Step by Step
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.
