Science:Math Exam Resources/Courses/MATH101/April 2014/Question 07 (b)/Solution 1

First, if we evaluate a few terms we see that $\cos(\pi )=-1$, $\cos(2\pi )=1$, $\cos(3\pi )=-1$, $\cdots$, $\cos(n\pi )=(-1)^{n}$ and therefore the terms alternate. Thus, the given series is alternating series,

${\begin{aligned}\sum _{n=1}^{\infty }(-1)^{n}{\frac {n}{2^{n}}}.\end{aligned}}$

Since $\lim _{n\to \infty }{\frac {n}{2^{n}}}=0$ and $a_{n}$ is decreasing for $n\geq 3$, by the alternating series test, the series $\sum _{n=1}^{\infty }{\frac {n\cos(n\pi )}{2^{n}}}$ converges.