,

$$\begin{eqnarray*} a_{1} & = & \sin\left(\frac{\pi}{2}-\pi\right)+\cos\pi=\left(-1\right)+\left(-1\right)=-2\\ a_{2} & = & \sin\left(\frac{\pi}{2}-2\pi\right)+\cos2\pi=\left(1\right)+\left(1\right)=2\\ a_{3} & = & \sin\left(\frac{\pi}{2}-3\pi\right)+\cos3\pi=\left(-1\right)+\left(-1\right)=-2\\ & \vdots \end{eqnarray*}$$

ซึ่งจะเห็นว่า

$$\left\{a_n\right\}=\left\{-2,2,-2,\cdots\right\}$$

$$\begin{eqnarray*} b_{1} & = & 10\cos\left(2\pi-\frac{\pi}{3}\right)=10\left(\frac{1}{2}\right)=5\\ b_{2} & = & 10\cos\left(4\pi-\frac{\pi}{3}\right)=10\left(\frac{1}{2}\right)=5\\ b_{3} & = & 10\cos\left(6\pi-\frac{\pi}{3}\right)=10\left(\frac{1}{2}\right)=5\\ & \vdots \end{eqnarray*}$$

จึงได้ว่า 

$$\left\{b_n\right\}=\left\{5,5,5,\cdots\right\}$$

,

จะเห็นว่าเป็นอนุกรมเรขาคณิตที่มี $r=-\frac25$ จึงใช้สูตรคำนวณผลรวมอนันต์ของอนุกรมเรขาคณิต $s_\infty = \frac{a_1}{1-r}$

$$\begin{eqnarray*} \frac{a_{1}}{b_{1}}+\left(\frac{a_{2}}{b_{2}}\right)^{2}+\left(\frac{a_{3}}{b_{3}}\right)^{3}+\cdots& = & \left(-\frac{2}{5}\right)+\left(-\frac{2}{5}\right)^{2}+\left(-\frac{2}{5}\right)^{3}+\cdots\\ & = & \frac{-\frac{2}{5}}{1-\left(-\frac{2}{5}\right)}\\ & = & -\frac{2}{7} \end{eqnarray*}$$