009C Sample Midterm 3, Problem 4

Test the series for convergence or divergence.

(a) (6 points)      Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\sin\frac{\pi}{n}.}

(b) (6 points)      ${\displaystyle \sum _{n=1}^{\infty }}\,(-1)^{n}\cos {\frac {\pi }{n}}.$

Foundations:
For $n\geq 2$, both sine and cosine of ${\frac {\pi }{n}}$ are strictly nonnegative. Thus, these series are alternating, and we can apply the
Alternating Series Test: If a series $\sum _{k=1}^{\infty }a_{k}$ is
Alternating in sign, and
$\lim _{k\rightarrow 0}|a_{k}|=0,$
then the series is convergent.
Note that if the series does not converge to zero, we must claim it diverges by the
Divergence Test: If ${\displaystyle \lim _{k\rightarrow \infty }a_{k}\neq 0,}$ then the series/sum $\sum _{k=0}^{\infty }a_{k}$ diverges.
In the case of an alternating series, such as the two listed for this problem, we can choose to show it does not converge to zero absolutely.

 Solution:

(a):
Here, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}{\displaystyle \lim _{n\rightarrow \infty }\left|(-1)^{n}\sin \left({\frac {\pi }{n}}\right)\right|}&=&\left|\sin \left({\displaystyle \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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}{\displaystyle \lim _{n\rightarrow \infty }\left|(-1)^{n}\cos \left({\frac {\pi }{n}}\right)\right|}&=&\left|\cos \left({\displaystyle \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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