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	<title>009C Sample Final 1, Problem 5 - Revision history</title>
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		<title>MathAdmin: Created page with &quot;&lt;span class=&quot;exam&quot;&gt; Let  ::::::&lt;math&gt;f(x)=\sum_{n=1}^{\infty} nx^n&lt;/math&gt;  ::&lt;span class=&quot;exam&quot;&gt;a) Find the radius of convergence of the power series.  ::&lt;span class=&quot;exam&quot;&gt;b)...&quot;</title>
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		<updated>2016-04-19T01:23:23Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt; Let  ::::::&amp;lt;math&amp;gt;f(x)=\sum_{n=1}^{\infty} nx^n&amp;lt;/math&amp;gt;  ::&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;a) Find the radius of convergence of the power series.  ::&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;b)...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt; Let&lt;br /&gt;
&lt;br /&gt;
::::::&amp;lt;math&amp;gt;f(x)=\sum_{n=1}^{\infty} nx^n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;a) Find the radius of convergence of the power series.&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;b) Determine the interval of convergence of the power series.&lt;br /&gt;
&lt;br /&gt;
::&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;c) Obtain an explicit formula for the function &amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;f(x)&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Foundations: &amp;amp;nbsp; &lt;br /&gt;
|-&lt;br /&gt;
|Recall:&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::'''1. Ratio Test''' Let &amp;lt;math style=&amp;quot;vertical-align: -7px&amp;quot;&amp;gt;\sum a_n&amp;lt;/math&amp;gt; be a series and &amp;lt;math&amp;gt;L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.&amp;lt;/math&amp;gt; Then,&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
:::If &amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;L&amp;lt;1,&amp;lt;/math&amp;gt; the series is absolutely convergent. &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
:::If &amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;L&amp;gt;1,&amp;lt;/math&amp;gt; the series is divergent.&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
:::If &amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;L=1,&amp;lt;/math&amp;gt; the test is inconclusive.&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
:::for convergence since the Ratio Test is inconclusive when &amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;L=1.&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
'''Solution:'''&lt;br /&gt;
&lt;br /&gt;
'''(a)'''&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 1: &amp;amp;nbsp; &lt;br /&gt;
|-&lt;br /&gt;
|To find the radius of convergence, we use the ratio test. We have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::&amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} &amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)x^{n+1}}{nx^n}}\bigg|\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{n+1}{n}x\bigg|}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{|x|\lim_{n \rightarrow \infty}\frac{n+1}{n}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{|x|.}\\&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 2: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|Thus, we have &amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;|x|&amp;lt;1&amp;lt;/math&amp;gt; and the radius of convergence of this series is &amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;1.&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
'''(b)'''&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 1: &amp;amp;nbsp; &lt;br /&gt;
|-&lt;br /&gt;
|From part (a), we know the series converges inside the interval &amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;(-1,1).&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Now, we need to check the endpoints of the interval for convergence.&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 2: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|For &amp;lt;math style=&amp;quot;vertical-align: -2px&amp;quot;&amp;gt;x=1,&amp;lt;/math&amp;gt; the series becomes &amp;lt;math&amp;gt;\sum_{n=1}^{\infty}n,&amp;lt;/math&amp;gt; which diverges by the Divergence Test.&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 3: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|For &amp;lt;math style=&amp;quot;vertical-align: -2px&amp;quot;&amp;gt;x=-1,&amp;lt;/math&amp;gt; the series becomes &amp;lt;math&amp;gt;\sum_{n=1}^{\infty}(-1)^n n,&amp;lt;/math&amp;gt; which diverges by the Divergence Test.&lt;br /&gt;
|-&lt;br /&gt;
|Thus, the interval of convergence is &amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;(-1,1).&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
'''(c)'''&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 1: &amp;amp;nbsp; &lt;br /&gt;
|-&lt;br /&gt;
|Recall that we have the geometric series formula &amp;lt;math&amp;gt;\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n&amp;lt;/math&amp;gt; for &amp;lt;math&amp;gt;|x|&amp;lt;1.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Now, we take the derivative of both sides of the last equation to get&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::&amp;lt;math&amp;gt;\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Step 2: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|Now, we multiply the last equation in Step 1 by &amp;lt;math style=&amp;quot;vertical-align: 0px&amp;quot;&amp;gt;x.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|So, we have &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::&amp;lt;math&amp;gt;\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n}=f(x).&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Thus, &amp;lt;math&amp;gt;f(x)=\frac{x}{(1-x)^2}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
{| class=&amp;quot;mw-collapsible mw-collapsed&amp;quot; style = &amp;quot;text-align:left;&amp;quot;&lt;br /&gt;
!Final Answer: &amp;amp;nbsp; &lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp;&amp;amp;nbsp; '''(a)''' &amp;lt;math style=&amp;quot;vertical-align: -3px&amp;quot;&amp;gt;1&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp;&amp;amp;nbsp; '''(b)''' &amp;lt;math style=&amp;quot;vertical-align: -3px&amp;quot;&amp;gt;(-1,1)&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp;&amp;amp;nbsp; '''(c)''' &amp;lt;math style=&amp;quot;vertical-align: -18px&amp;quot;&amp;gt;f(x)=\frac{x}{(1-x)^2}&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
[[009C_Sample_Final_1|'''&amp;lt;u&amp;gt;Return to Sample Exam&amp;lt;/u&amp;gt;''']]&lt;/div&gt;</summary>
		<author><name>MathAdmin</name></author>
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