009C Sample Midterm 1, Problem 4 - Revision history

Kayla Murray at 05:55, 12 November 2017

2017-11-12T05:55:03Z

Kayla Murray at 01:05, 5 November 2017

2017-11-05T01:05:59Z

Kayla Murray at 17:16, 18 March 2017

2017-03-18T17:16:18Z

Kayla Murray at 01:55, 27 February 2017

2017-02-27T01:55:09Z

Kayla Murray at 00:17, 19 February 2017

2017-02-19T00:17:24Z

Kayla Murray at 22:35, 15 February 2017

2017-02-15T22:35:49Z

Kayla Murray at 22:32, 15 February 2017

2017-02-15T22:32:01Z

Kayla Murray at 16:54, 15 February 2017

2017-02-15T16:54:51Z

Kayla Murray at 23:22, 12 February 2017

2017-02-12T23:22:57Z

Kayla Murray at 16:43, 5 February 2017

2017-02-05T16:43:56Z

@@ Line 7: / Line 7: @@
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 1, Problem 4 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 6: / Line 6: @@
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+(insert picture of handwritten solution)
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; converges
+|&nbsp; &nbsp; &nbsp; &nbsp; converges (by the Direct Comparison Test)
 |-
+|
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 |'''Direct Comparison Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> and <math>\{b_n\}</math> be positive sequences where <math style="vertical-align: -3px">a_n\le b_n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; and &nbsp;<math>\{b_n\}</math>&nbsp; be positive sequences where &nbsp;<math style="vertical-align: -3px">a_n\le b_n</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; for all <math style="vertical-align: -3px">n\ge N</math> for some <math style="vertical-align: -3px">N\ge 1.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; for all &nbsp;<math style="vertical-align: -3px">n\ge N</math>&nbsp; for some &nbsp;<math style="vertical-align: -3px">N\ge 1.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; '''1.''' If <math>\sum_{n=1}^\infty b_n</math> converges, then <math>\sum_{n=1}^\infty a_n</math> converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; '''1.''' If &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges, then &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; converges.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; '''2.''' If <math>\sum_{n=1}^\infty a_n</math> diverges, then <math>\sum_{n=1}^\infty b_n</math> diverges.
+|&nbsp; &nbsp; &nbsp; &nbsp; '''2.''' If &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; diverges, then &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges.
 |}
@@ Line 30: / Line 30: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n^23^n}>0</math>
 |-
-|for all <math style="vertical-align: -3px">n\ge 1.</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 <math style="vertical-align: -14px">a_n=\frac{1}{n^23^n}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">a_n=\frac{1}{n^23^n}.</math>
 |}
@@ Line 40: / Line 40: @@
 !Step 2: &nbsp;
 |-
-|Let <math style="vertical-align: -14px">b_n=\frac{1}{n^2}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n^2}.</math>
 |-
 |We want to compare the series in this problem with
@@ Line 46: / Line 46: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n^2}.</math>
 |-
-|This is a <math style="vertical-align: -4px">p</math>-series with <math style="vertical-align: -4px">p=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, <math>\sum_{n=1}^\infty b_n</math> converges.
+|Hence, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges.
 |}
@@ Line 54: / Line 54: @@
 !Step 3: &nbsp;
 |-
-|Also, we have <math style="vertical-align: -4px">a_n<b_n</math> since
+|Also, we have &nbsp;<math style="vertical-align: -4px">a_n<b_n</math>&nbsp; since
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n^23^n}<\frac{1}{n^2}</math>
 |-
-|for all <math style="vertical-align: -3px">n\ge 1.</math>
+|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 |-
-|Therefore, the series <math>\sum_{n=1}^\infty a_n</math> converges
+|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; converges
 |-
 |by the Direct Comparison Test.

← Older revision		Revision as of 00:17, 19 February 2017
Line 3:		Line 3:
	<span class="exam"> Be sure to justify your answers!		<span class="exam"> Be sure to justify your answers!

−	~~::::~~::<math>\sum_{n=1}^\infty \frac{1}{n^23^n}</math>	+	::<math>\sum_{n=1}^\infty \frac{1}{n^23^n}</math>

@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
+|'''Direct Comparison Test'''
 |-
+|
 |-
+|
 |}
 '''Solution:'''
@@ Line 23: / Line 22: @@
 !Step 1: &nbsp;
 |-
+|First, we note that
 |-
+|Let <math>a_n=\frac{1}{n^23^n}.</math>
 |}
@@ Line 31: / Line 36: @@
 !Step 2: &nbsp;
 |-
+|Let <math>b_n=\frac{1}{n^2}.</math>
 |-
+|by the Direct Comparison Test.
 |}
@@ Line 40: / Line 65: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; converges
 |-
+|
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">
+<span class="exam">Determine the convergence or divergence of the following series.
-::<span class="exam">a)
+<span class="exam"> Be sure to justify your answers!
-::<span class="exam">b)
@@ Line 19: / Line 20: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
@@ Line 36: / Line 36: @@
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
+!Final Answer: &nbsp;
 |-
+|
 |-
+|
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]