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:
Direct Comparison Test
Let  $\{a_{n}\}$  and  $\{b_{n}\}$  be positive sequences where  $a_{n}\leq b_{n}$
for all  $n\geq N$  for some  $N\geq 1.$
1. If  $\sum _{n=1}^{\infty }b_{n}$  converges, then  $\sum _{n=1}^{\infty }a_{n}$  converges.
2. If  $\sum _{n=1}^{\infty }a_{n}$  diverges, then  $\sum _{n=1}^{\infty }b_{n}$  diverges.

Solution:

Step 1:
First, we note that
${\frac {1}{n^{2}3^{n}}}>0$
for all  $n\geq 1.$
This means that we can use a comparison test on this series.
Let  $a_{n}={\frac {1}{n^{2}3^{n}}}.$

Step 2:
Let  $b_{n}={\frac {1}{n^{2}}}.$
We want to compare the series in this problem with
$\sum _{n=1}^{\infty }b_{n}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}.$
This is a  $p$-series with  $p=2.$
Hence,  $\sum _{n=1}^{\infty }b_{n}$  converges.

Step 3:
Also, we have  $a_{n}<b_{n}$  since
${\frac {1}{n^{2}3^{n}}}<{\frac {1}{n^{2}}}$
for all  $n\geq 1.$
Therefore, the series  $\sum _{n=1}^{\infty }a_{n}$  converges
by the Direct Comparison Test.

Final Answer:
converges (by the Direct Comparison Test)

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