009C Sample Midterm 2, Problem 3 - Revision history

Kayla Murray at 19:35, 12 November 2017

2017-11-12T19:35:27Z

Kayla Murray: Replaced content with "Determine convergence or divergence: (a) $\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}$ (b) &nbs..."

2017-11-05T01:12:59Z

Replaced content with "<span class="exam">Determine convergence or divergence: <span class="exam">(a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math> <span class="exam">(b) &nbs..."

Kayla Murray at 17:38, 18 March 2017

2017-03-18T17:38:53Z

Kayla Murray at 02:08, 27 February 2017

2017-02-27T02:08:05Z

Kayla Murray at 19:07, 26 February 2017

2017-02-26T19:07:54Z

Kayla Murray at 00:19, 19 February 2017

2017-02-19T00:19:53Z

Kayla Murray at 00:15, 16 February 2017

2017-02-16T00:15:19Z

Kayla Murray at 19:29, 13 February 2017

2017-02-13T19:29:21Z

Kayla Murray at 19:06, 13 February 2017

2017-02-13T19:06:27Z

Kayla Murray at 19:05, 13 February 2017

2017-02-13T19:05:40Z

@@ Line 6: / Line 6: @@
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 55: / Line 55: @@
 |-
 |Let &nbsp;<math> b_n=\frac{1}{\sqrt{n}}.</math>
 |-
 |The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
@@ Line 155: / Line 161: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges (by the Alternating Series Test)
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; converges
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; converges (by the Ratio Test)
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Determine convergence or divergence:
-<span class="exam">(a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math>
+<span class="exam">(a) &nbsp;<math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math>
-<span class="exam">(b) <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math>
+<span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math>
@@ Line 11: / Line 11: @@
 |'''1.''' '''Alternating Series Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; be a positive, decreasing sequence where &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math>\sum_{n=1}^\infty (-1)^na_n</math>&nbsp; and &nbsp;<math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; converge.
@@ Line 19: / Line 19: @@
 |'''2.''' '''Ratio Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -7px">\sum a_n</math>&nbsp; be a series and &nbsp;<math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Then,
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 |-
 |'''3.''' If a series absolutely converges, then it also converges.
@@ Line 54: / Line 54: @@
 |We notice that the series is alternating.
 |-
-|Let <math> b_n=\frac{1}{\sqrt{n}}.</math>
+|Let &nbsp;<math> b_n=\frac{1}{\sqrt{n}}.</math>
 |-
-|The sequence <math>\{b_n\}</math> is decreasing since
+|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{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 66: / Line 66: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
 |-
-|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges by the Alternating Series Test.
+|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>&nbsp; converges by the Alternating Series Test.
 |}
@@ Line 92: / Line 92: @@
 !Step 2: &nbsp;
 |-
-|Now, we need to calculate <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
+|Now, we need to calculate &nbsp;<math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
 |-
-|Let <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
+|Let &nbsp;<math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
 |-
 |Then, taking the natural log of both sides, we get
@@ Line 111: / Line 111: @@
 |since we can interchange limits and continuous functions.
 |-
-|Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
+|Now, this limit has the form &nbsp;<math style="vertical-align: -13px">\frac{0}{0}.</math>
 |-
 |Hence, we can use L'Hopital's Rule to calculate this limit.
@@ Line 138: / Line 138: @@
 !Step 4: &nbsp;
 |-
-|Since <math>\ln y=-1,</math> we know
+|Since &nbsp;<math>\ln y=-1,</math>&nbsp; we know
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{-1}.</math>
@@ Line 146: / Line 146: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math>
 |-
-|Since <math style="vertical-align: -13px">\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test.
+|Since &nbsp;<math style="vertical-align: -13px">\frac{2}{e}<1,</math>&nbsp; the series is absolutely convergent by the Ratio Test.
 |-
 |Therefore, the series converges.

@@ Line 36: / Line 36: @@
+|
 |}
 '''Solution:'''
@@ Line 149: / Line 150: @@
 |Therefore, the series converges.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"