Test the series for convergence or divergence.
 (a) (6 points) $\sum _{n=1}^{\infty }}\,(1)^{n}\sin {\frac {\pi }{n}}.$
 (b) (6 points) $\sum _{n=1}^{\infty }}\,(1)^{n}\cos {\frac {\pi }{n}}.$
Solution:
(a):

Here, we have

 ${\begin{array}{rcl} \lim _{n\rightarrow \infty }\left(1)^{n}\sin \left({\frac {\pi }{n}}\right)\right}&=&\left\sin \left( \lim _{n\rightarrow \infty }{\frac {\pi }{n}}}\right)\right\\\\&=&0,\end{array}}$

as we can pass the limit through absolute value and sine because both are continuous functions. Since the series alternates for $n\geq 2$, the series is convergent by the alternating series test.

(b):

In this case, we find

 ${\begin{array}{rcl} \lim _{n\rightarrow \infty }\left(1)^{n}\cos \left({\frac {\pi }{n}}\right)\right}&=&\left\cos \left( \lim _{n\rightarrow \infty }{\frac {\pi }{n}}}\right)\right\\\\&=&\cos(0)\\\\&=&1,\end{array}}$

where we can again pass the limit through absolute value and cosine because both are continuous. Since the terms do not converge to zero absolutely, they do not converge to zero. Thus, by the divergence test, the series is divergent.

Final Answer:

The series for (a) is convergent, while the series (b) is divergent.

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