009C Sample Midterm 1, Problem 3 - Revision history

Kayla Murray at 05:54, 12 November 2017

2017-11-12T05:54:03Z

Kayla Murray at 01:02, 5 November 2017

2017-11-05T01:02:07Z

Kayla Murray at 17:16, 18 March 2017

2017-03-18T17:16:43Z

Kayla Murray at 17:14, 18 March 2017

2017-03-18T17:14:23Z

Kayla Murray at 01:53, 27 February 2017

2017-02-27T01:53:00Z

Kayla Murray at 00:17, 19 February 2017

2017-02-19T00:17:10Z

Kayla Murray at 16:47, 15 February 2017

2017-02-15T16:47:07Z

Kayla Murray at 16:44, 15 February 2017

2017-02-15T16:44:07Z

Kayla Murray at 17:21, 14 February 2017

2017-02-14T17:21:26Z

Kayla Murray at 16:39, 14 February 2017

2017-02-14T16:39:38Z

@@ Line 9: / Line 9: @@
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
-{| 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 99: / Line 99: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Conditionally convergent (by the p-test and the Alternating Series Test)
+|&nbsp; &nbsp; &nbsp; &nbsp; conditionally convergent (by the p-test and the Alternating Series Test)
 |-
+|
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 65: / Line 65: @@
 |-
 |Let &nbsp;<math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
 |-
 |The sequence &nbsp;<math style="vertical-align: -4px">\{b_n\}</math>&nbsp; is decreasing since
@@ Line 93: / Line 99: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Conditionally convergent
+|&nbsp; &nbsp; &nbsp; &nbsp; Conditionally convergent (by the p-test and the Alternating Series Test)
 |-
+|
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' A series <math>\sum a_n</math> is '''absolutely convergent''' if
+|'''1.''' A series &nbsp;<math>\sum a_n</math>&nbsp; is '''absolutely convergent''' if
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; the series <math>\sum |a_n|</math> converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; the series &nbsp;<math>\sum |a_n|</math>&nbsp; converges.
 |-
-|'''2.''' A series <math>\sum a_n</math> is '''conditionally convergent''' if
+|'''2.''' A series &nbsp;<math>\sum a_n</math>&nbsp; is '''conditionally convergent''' if
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; the series <math>\sum |a_n|</math> diverges and the series <math>\sum a_n</math> converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; the series &nbsp;<math>\sum |a_n|</math>&nbsp; diverges and the series &nbsp;<math>\sum a_n</math>&nbsp; converges.
 |}
@@ Line 28: / Line 28: @@
 |First, we take the absolute value of the terms in the original series.
 |-
-|Let <math style="vertical-align: -14px">a_n=\frac{(-1)^n}{n}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">a_n=\frac{(-1)^n}{n}.</math>
 |-
 |Therefore,
@@ Line 42: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>).
+|This series is the harmonic series (or &nbsp;<math style="vertical-align: -5px">p</math>-series with &nbsp;<math style="vertical-align: -5px">p=1</math>&nbsp;).
 |-
 |Thus, it diverges. Hence, the series
@@ Line 64: / Line 64: @@
 |we notice that this series is alternating.
 |-
-|Let <math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
+|Let &nbsp;<math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
 |-
-|The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since
+|The sequence &nbsp;<math style="vertical-align: -4px">\{b_n\}</math>&nbsp; is decreasing since
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}<\frac{1}{n}</math>
 |-
-|for all <math style="vertical-align: -3px">n\ge 1.</math>
+|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 |-
 |Also,
@@ Line 76: / Line 76: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>
 |-
-|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges
+|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges
 |-
 |by the Alternating Series Test.
@@ Line 84: / Line 84: @@
 !Step 4: &nbsp;
 |-
-|Since the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; is not absolutely convergent but convergent,
+|Since the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; is not absolutely convergent but convergent,
 |-
 |this series is conditionally convergent.

← Older revision		Revision as of 00:17, 19 February 2017
Line 5:		Line 5:
	<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}{n}</math>	+	::<math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>

@@ Line 44: / Line 44: @@
 |This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>).
 |-
-|So, it diverges. Hence, the series
+|Thus, it diverges. Hence, the series
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
@@ Line 58: / Line 58: @@
 |if it conditionally converges.
 |-
-|For <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
+|For
 |-
 |we notice that this series is alternating.
@@ Line 74: / Line 76: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>
 |-
-|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges by the Alternating Series Test.
+|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges
 |}

← Older revision		Revision as of 16:44, 15 February 2017
Line 1:		Line 1:
−	<span class="exam">Determine whether the following series converges absolutely, conditionally or whether it diverges.	+	<span class="exam"> Determine whether the following series converges absolutely,
		+
		+	<span class="exam"> conditionally or whether it diverges.

	<span class="exam"> Be sure to justify your answers!		<span class="exam"> Be sure to justify your answers!