Difference between revisions of "009C Sample Midterm 2, Problem 5"

Revision as of 12:55, 24 April 2017

If $\sum _{n=0}^{\infty }c_{n}x^{n}$ converges, does it follow that the following series converges?

(a) $\sum _{n=0}^{\infty }c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}$

(b) $\sum _{n=0}^{\infty }c_{n}(-x)^{n}$

Foundations:
Ratio Test
Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum a_{n}} be a series and $L=\lim _{n\rightarrow \infty }{\bigg \|}{\frac {a_{n+1}}{a_{n}}}{\bigg \|}.$
Then,
If $L<1,$ the series is absolutely convergent.
If $L>1,$ the series is divergent.
If $L=1,$ the test is inconclusive.

Solution:

(a)

Step 1:
Assume that the power series $\sum _{n=0}^{\infty }c_{n}x^{n}$ converges.
Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R} be the radius of convergence of this power series.
We can use the Ratio Test to find $R.$
Using the Ratio Test, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg \|}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}x}{c_{n}}}{\bigg \|}}\\&&\\&=&\displaystyle {\|x\|\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}}
Since the radius of convergence of the series $\sum _{n=0}^{\infty }c_{n}x^{n}$ is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R,} we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R={\frac {1}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}.}

Step 2:
Now, we use the Ratio Test to find the radius of convergence of the series Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sum _{n=0}^{\infty }c_{n}{\bigg (}{\frac {x}{2}}{\bigg )}^{n}.}
Using the Ratio Test, we have
${\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}2^{n}x^{n+1}}{c_{n}2^{n+1}x^{n}}}{\bigg \|}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}x}{2c_{n}}}{\bigg \|}}\\&&\\&=&\displaystyle {{\frac {\|x\|}{2}}\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}$
Hence, the radius of convergence of this power series is
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {2}{\displaystyle {\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}}}}=2R.}
Therefore, this power series converges.

(b)

Step 1:
Assume that the power series $\sum _{n=0}^{\infty }c_{n}x^{n}$ converges.
Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle R} be the radius of convergence of this power series.
We can use the Ratio Test to find $R.$
Using the Ratio Test, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}x^{n+1}}{c_{n}x^{n}}}{\bigg \|}}&=&\displaystyle {\lim _{n\rightarrow \infty }{\bigg \|}{\frac {c_{n+1}x}{c_{n}}}{\bigg \|}}\\&&\\&=&\displaystyle {\|x\|\lim _{n\rightarrow \infty }{\frac {c_{n+1}}{c_{n}}}.}\end{array}}}
Since the radius of convergence of the series $\sum _{n=0}^{\infty }c_{n}x^{n}$ is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R,} we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.}

Step 2:
Now, we use the Ratio Test to find the radius of convergence of the series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty c_n(-x)^n .}
Using the Ratio Test, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} \bigg\| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg\|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\|\frac{c_{n+1}(-x)}{c_n}\bigg\|}\\ &&\\ & = & \displaystyle{\|x\|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.} \end{array}}
Hence, the radius of convergence of this power series is
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.}
Therefore, this power series converges.

Final Answer:
(a) converges
(b) converges

Return to Sample Exam

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|If a power series converges, then it has a nonempty interval of convergence.
+|'''Ratio Test'''
+|-
+|&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 &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
+|-
+|
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
+|-
+|
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 |}
@@ Line 24: / Line 37: @@
 |Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 |-
-|So, the power series
+|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+|
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
+&&\\
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+\end{array}</math>
+|-
+|Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
 |}
@@ Line 34: / Line 56: @@
 !Step 2: &nbsp;
 |-
-|Let &nbsp;<math style="vertical-align: -5px">a\in (-2R,2R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math>&nbsp;
+|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|&nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math>&nbsp; converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}2^nx^{n+1}}{c_n2^{n+1}x^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{2c_n}\bigg|}\\
+&&\\
+& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+\end{array}</math>
 |-
-|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R),</math>&nbsp;
+|Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.</math>
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R).</math>&nbsp;
+|Therefore, this power series converges.
 |}
@@ Line 55: / Line 81: @@
 |Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 |-
-|So, the power series
+|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
+|-
+|Using the Ratio Test, we have
+|-
+|
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
+&&\\
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+\end{array}</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;
+|Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
 |}
@@ Line 65: / Line 100: @@
 !Step 2: &nbsp;
 |-
-|Let &nbsp;<math style="vertical-align: -5px">a\in (-R,R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">-a \in (-R,R).</math>&nbsp;
+|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n(-x)^n .</math>
 |-
-|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|&nbsp;<math>\sum_{n=0}^\infty c_n(-a)^n</math>&nbsp; converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}(-x)}{c_n}\bigg|}\\
+&&\\
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+\end{array}</math>
 |-
-|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n(-x)^n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.</math>
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+|Therefore, this power series converges.
 |}

Difference between revisions of "009C Sample Midterm 2, Problem 5"

Revision as of 12:55, 24 April 2017

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