Difference between revisions of "009C Sample Final 3, Problem 3"

Latest revision as of 18:43, 2 December 2017

Test if the following series converges or diverges. Give reasons and clearly state if you are using any standard test.

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 \sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}}

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Detailed Solution

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@@ Line 3: / Line 3: @@
 ::<math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 3 Solution|'''<u>Solution</u>''']]
-|-
-|'''Limit Comparison Test'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; and &nbsp;<math>\{b_n\}</math>&nbsp; be positive sequences.
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -16px">\lim_{n\rightarrow \infty} \frac{a_n}{b_n}=L,</math>&nbsp; where &nbsp;<math style="vertical-align: -1px">L</math> &nbsp;is a positive real number,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -20px">\sum_{n=1}^\infty a_n</math>&nbsp; and &nbsp;<math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>&nbsp; either both converge or both diverge.
-|}
-'''Solution:'''
+[[009C Sample Final 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we note that
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{n^3+7n}{\sqrt{1+n^{10}}}>0</math>
-|-
-|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
-|-
-|This means that we can use a comparison test on this series.
-|-
-|Let &nbsp;<math style="vertical-align: -19px">a_n=\frac{n^3+7n}{\sqrt{1+n^{10}}}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n^2}.</math>
-|-
-|We want to compare the series in this problem with
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n^2}.</math>
-|-
-|This is a &nbsp;<math style="vertical-align: -4px">p</math>-series with &nbsp;<math style="vertical-align: -4px">p=2.</math>
-|-
-|Hence, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{n^3+7n}{\sqrt{1+n^{10}}})}{(\frac{1}{n^2})}}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n^5+7n^3}{\sqrt{1+n^{10}}}}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n^5+7n^3}{\sqrt{1+n^{10}}} \bigg(\frac{\frac{1}{n^5}}{\frac{1}{n^5}}\bigg)}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1+\frac{7}{n^4}}{\sqrt{\frac{1}{n^{10}}+1}}}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|-
-|Therefore, the series
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^{\infty} \frac{n^3+7n}{\sqrt{1+n^{10}}}</math>
-|-
-|converges by the Limit Comparison Test.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; converges
-|}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 3, Problem 3"

Latest revision as of 18:43, 2 December 2017

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