009C Sample Midterm 3, Problem 3 - Revision history

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MathAdmin at 20:59, 26 April 2015

2015-04-26T20:59:49Z

MathAdmin: Created page with "Test if each the following series converges or diverges. Give reasons and clearly state if you are using any standard test. ::(a) (6 poi..."

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Created page with "Test if each the following series converges or diverges. Give reasons and clearly state if you are using any standard test. ::(a) (6 poi..."

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

::(a) (6 points)      <math>{\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+1)!}.</math>
 

::(b) (6 points)      <math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|Most of the time, if there are factorials in the terms of a series, you would use the
|-
|'''Ratio Test.''' Let <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> be a series. Then:
|-
|
:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, the series is absolutely convergent (and therefore convergent).
|-
|
:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -83%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty</math>, the series is divergent.
|-
|
:*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.
|-
|This works well, as factorials cancel out many terms. For example,
|-
|
::<math>\frac{(n+1)!}{n!}\,=\,n+1.</math>
|-
|On the other hand, something built mainly out of powers of <math style="vertical-align: 0%">n</math> may work well with the
|-
|'''Limit Comparison Test.''' Suppose <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> are series with positive terms. If <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: -5%">0<c<\infty</math>, then either both series converge, or both series diverge.
|-
|In the case of a series built mainly out of powers, you would choose to compare it to a ''p''-series.
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a):  
|-
|As mentioned in Foundations, we should use the ratio test. Note that
|-
|
::<math style="vertical-align: 0%">\begin{array}{rcl}
\displaystyle {\left|\frac{a_{n+1}}{a_{n}}\right|} & = & \left|\frac{\frac{(n+1)!}{(3(n+1)+1)!}}{\frac{n!}{(3n+1)!}}\right|\\
\\
& = & \displaystyle {\frac{(n+1)!}{n!}\cdot\frac{(3n+1)!}{(3n+4)!}}\\
\\
& = & \displaystyle {\frac{n+1}{(3n+2)(3n+3)(3n+4)}}.
\end{array}</math>
|-
|Thus,
|-
|
::<math style="vertical-align: 0%">\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|\ =\ 0\ <\ 1,</math>
|-
|so by the ratio test the series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b):  
|-
|Here, we can use the limit comparison test. Let <math style="vertical-align: -62%">a_{n}=\frac{\sqrt{n}}{n^{2}-3}</math>, and let <math style="vertical-align: -62%">b_{n}=\frac{1}{n^{3/2}}.</math> Notice that the terms of <math style="vertical-align: -10%">a_{n}</math> are all positive, and
|-
|
::<math style="vertical-align: 0%">\begin{array}{rcl}
\displaystyle {\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}} & = & \displaystyle {\lim_{n\rightarrow\infty}\frac{\frac{\sqrt{n}}{n^{2}-3}}{\frac{1}{n^{3/2}}}}\\
\\
& = & \displaystyle {\lim_{n\rightarrow\infty}\frac{n^{2}}{n^{2}-3}}\\
\\
& = & 1\,\,<\,\, \infty.
\end{array}</math>

Since <math style="vertical-align: -90%">\sum_{n=2}^{\infty}\frac{1}{n^{3/2}}</math>  is a p-series with <math style="vertical-align: -20%">p>1,</math>
it is convergent. By the limit comparison test, <math style="vertical-align: -90%">\sum_{n=2}^{\infty}\frac{\sqrt{n}}{n^{2}-3}</math>
 is convergent.

|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Test if each the following series converges or diverges. Give reasons
+<span class="exam">Test if each the following series converges or diverges.
-::<span class="exam">(a) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+1)!}.</math>
+<span class="exam">Give reasons and clearly state if you are using any standard test.
 <br>
-::<span class="exam">(b) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}.</math>
+<span class="exam">(b) &nbsp;<math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}</math>
+<hr>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 76: / Line 76: @@
 it is convergent. By the limit comparison test, <math style="vertical-align: -90%">\sum_{n=2}^{\infty}\frac{\sqrt{n}}{n^{2}-3}</math>
 &thinsp;is convergent.
 |}
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]