009C Sample Midterm 3, Problem 4 Detailed Solution

Test the series for convergence or divergence.

(a)  ${\displaystyle \sum _{n=1}^{\infty }}\,(-1)^{n}\sin {\frac {\pi }{n}}$

(b)  ${\displaystyle \sum _{n=1}^{\infty }}\,(-1)^{n}\cos {\frac {\pi }{n}}$

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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.

Final Answer:
(a)     converges (by the Alternating Series Test)
(b)     diverges (by the Divergence Test)

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