009C Sample Midterm 3, Problem 4 Detailed Solution

Test the series for convergence or divergence.

(a)  $\sum _{n=1}^{\infty }}\,(-1)^{n}\sin {\frac {\pi }{n}}$ (b)  $\sum _{n=1}^{\infty }}\,(-1)^{n}\cos {\frac {\pi }{n}}$ Background Information:
Alternating Series Test
Let  $\{a_{n}\}$ be a positive, decreasing sequence where  $\lim _{n\rightarrow \infty }a_{n}=0.$ Then,  $\sum _{n=1}^{\infty }(-1)^{n}a_{n}$ and  $\sum _{n=1}^{\infty }(-1)^{n+1}a_{n}$ converge.

Solution:

(a)

Step 1:
First, we note that
$\sin {\bigg (}{\frac {\pi }{n}}{\bigg )}>0$ for all  $n\geq 1.$ So, the series
$\sum _{n=1}^{\infty }(-1)^{n}\sin {\bigg (}{\frac {\pi }{n}}{\bigg )}$ is alternating.
Let  $b_{n}=\sin {\bigg (}{\frac {\pi }{n}}{\bigg )}.$ Step 2:
The sequence  $\{b_{n}\}$ is decreasing since
$\sin {\bigg (}{\frac {\pi }{n+1}}{\bigg )}<\sin {\bigg (}{\frac {\pi }{n}}{\bigg )}$ for all  $n\geq 2.$ Also,

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }b_{n}}&=&\displaystyle {\lim _{n\rightarrow \infty }\sin {\bigg (}{\frac {\pi }{n}}{\bigg )}}\\&&\\&=&\displaystyle {\sin(0)}\\&&\\&=&\displaystyle {0.}\end{array}}$ Therefore,
$\sum _{n=1}^{\infty }(-1)^{n}\sin {\frac {\pi }{n}}$ converges by the Alternating Series Test.

(b)

Step 1:
First, we note that
$\cos {\bigg (}{\frac {\pi }{n}}{\bigg )}>0$ for all  $n\geq 3.$ So, the series
$\sum _{n=1}^{\infty }(-1)^{n}\cos {\bigg (}{\frac {\pi }{n}}{\bigg )}$ is alternating.
Also, we have

${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }\cos {\bigg (}{\frac {\pi }{n}}{\bigg )}}&=&\displaystyle {\cos(0)}\\&&\\&=&\displaystyle {1.}\end{array}}$ Step 2:
Since  $\lim _{n\rightarrow \infty }\cos {\bigg (}{\frac {\pi }{n}}{\bigg )}\neq 0,$ we have
$\lim _{n\rightarrow \infty }(-1)^{n}\cos {\bigg (}{\frac {\pi }{n}}{\bigg )}=DNE.$ Therefore, the series diverges by the Divergence Test.