009C Sample Final 3, Problem 6 - Revision history

Kayla Murray at 23:21, 3 December 2017

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@@ Line 1: / Line 1: @@
 <span class="exam"> Consider the power series
-::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}</math>
+::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
 <span class="exam">(a) Find the radius of convergence of the above power series.

@@ Line 274: / Line 274: @@
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>f(x)=\ln(1+x)</math>
 |-
-|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; converges
+|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; converges (by the Direct Comparison Test)
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 13: / Line 13: @@
 ::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>
-<span class="exam">converge? If so, find its sum.
+<span class="exam">converge?
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 100: / Line 100: @@
 |-
 |Let &nbsp;<math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
 |-
 |The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since

@@ Line 205: / Line 205: @@
 \end{array}</math>
 |-
-|Since there is no constant term in the series <math style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},~C=0.</math>
+|Since there is no constant term in the series <math style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
 |-
 |Hence,

@@ Line 197: / Line 197: @@
 !Step 2: &nbsp;
 |-
-|Then,
+|Thus,
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 205: / Line 205: @@
 \end{array}</math>
 |-
-|Since there is no constant term in the series <math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
+|Since there is no constant term in the series <math style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},~C=0.</math>
 |-
 |Hence,

@@ Line 173: / Line 173: @@
 !Step 1: &nbsp;
 |-
+|Let
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
 |-
+|Then,
 |}
@@ Line 183: / Line 197: @@
 !Step 2: &nbsp;
 |-
+|Then,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(x)=\ln(1+x).</math>
 |}
@@ Line 240: / Line 266: @@
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>(-1,1]</math>
 |-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>f(x)=\ln(1+x)</math>
 |-
 |&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; converges
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 193: / Line 193: @@
 !Step 1: &nbsp;
 |-
+|First, we note that
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)3^{n+1}}>0</math>
 |-
+|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 |}
@@ Line 203: / Line 207: @@
 !Step 2: &nbsp;
 |-
+|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{3^{n+1}}.</math>
 |-
+|by the Direct Comparison Test.
 |}
@@ Line 218: / Line 242: @@
 |&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;
 |-
-|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; converges
 |}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 137: / Line 137: @@
 |This means that we can use the limit comparison test on this series.
 |-
-|Let &nbsp;<math style="vertical-align: -19px">a_n=\frac{1}{n+1}.</math>
+|Let &nbsp;<math style="vertical-align: -16px">a_n=\frac{1}{n+1}.</math>
 |-
 |Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
 |-
-|Then, &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges since it is the harmonic series.
+|Then, &nbsp;<math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>&nbsp; diverges since it is the harmonic series.
 |-
 |We have