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| | <span class="exam">Test the series for convergence or divergence. | | <span class="exam">Test the series for convergence or divergence. |
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| − | ::<span class="exam">(a) (6 points) <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\sin\frac{\pi}{n}.</math>
| + | <span class="exam">(a) <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\sin\frac{\pi}{n}</math> |
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| − | ::<span class="exam">(b) (6 points) <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\cos\frac{\pi}{n}.</math>
| + | <span class="exam">(b) <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\cos\frac{\pi}{n}</math> |
| | + | <hr> |
| | + | [[009C Sample Midterm 3, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | ! Foundations:
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| − | |-
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| − | |For <math>n\geq2</math>, both sine and cosine of <math>\frac{\pi}{n}</math> are strictly nonnegative. Thus, these series are alternating, and we can apply the
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| − | |-
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| − | |'''Alternating Series Test:''' If a series <math>\sum_{k=1}^{\infty} a_{k}</math> is
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| − | |-
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| − | :*Alternating in sign, and
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| − | |-
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| − | :*<math>\lim_{k\rightarrow 0}|a_{k}|=0,</math>
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| − | |-
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| − | |then the series is convergent.
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| − | |-
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| − | |Note that if the series does ''<u>not</u>'' converge to zero, we must claim it diverges by the
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| − | '''Divergence Test:''' If <math style="vertical-align: -65%">{\displaystyle \lim_{k\rightarrow\infty}a_{k}\neq0,}</math> then the series/sum <math style="vertical-align: -98%">\sum_{k=0}^{\infty}a_{k}</math> diverges.
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| − | |-
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| − | |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.
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| − | |}
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| − | '''Solution:'''
| + | [[009C Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a):
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| − | |-
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| − | |Here, we have
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| − | |-
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| − | ::<math>\begin{array}{rcl}
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| − | {\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|\\
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| − | \\
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| − | & = & 0,
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| − | \end{array} </math>
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| − | |-
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| − | |<br>as we can pass the limit through absolute value and sine because both are continuous functions. Since the series alternates for <math style="vertical-align: -15%">n\geq 2</math>, the series is convergent by the alternating series test.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(b):
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| − | |-
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| − | |In this case, we find
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| − | |-
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | {\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|\\
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| − | \\
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| − | & = & \cos(0)\\
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| − | \\
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| − | & = & 1,
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| − | \end{array}</math>
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| − | |-
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| − | |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.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | |The series for (a) is convergent, while the series (b) is divergent.
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| − | |}
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| | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
