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	<title>009C Sample Final 3, Problem 1 - Revision history</title>
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	<updated>2026-04-22T17:02:11Z</updated>
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		<title>MathAdmin: Created page with &quot;&lt;span class=&quot;exam&quot;&gt; Which of the following sequences &amp;nbsp;&lt;math style=&quot;vertical-align: -5px&quot;&gt;(a_n)_{n\ge 1}&lt;/math&gt;&amp;nbsp; converges? Which diverges? Give reasons for your answ...&quot;</title>
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		<updated>2017-03-19T18:38:27Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt; Which of the following sequences  &amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;(a_n)_{n\ge 1}&amp;lt;/math&amp;gt;  converges? Which diverges? Give reasons for your answ...&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; Which of the following sequences &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;(a_n)_{n\ge 1}&amp;lt;/math&amp;gt;&amp;amp;nbsp; converges? Which diverges? Give reasons for your answers!&lt;br /&gt;
&lt;br /&gt;
&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;(a) &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -15px&amp;quot;&amp;gt;a_n=\bigg(1+\frac{1}{2n}\bigg)^n&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;(b) &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -15px&amp;quot;&amp;gt;a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n&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;
|'''L'Hôpital's Rule''' &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; Suppose that &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -11px&amp;quot;&amp;gt;\lim_{x\rightarrow \infty} f(x)&amp;lt;/math&amp;gt;&amp;amp;nbsp; and &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -11px&amp;quot;&amp;gt;\lim_{x\rightarrow \infty} g(x)&amp;lt;/math&amp;gt;&amp;amp;nbsp; are both zero or both &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -1px&amp;quot;&amp;gt;\pm \infty .&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; If &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -19px&amp;quot;&amp;gt;\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}&amp;lt;/math&amp;gt;&amp;amp;nbsp; is finite or &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -4px&amp;quot;&amp;gt;\pm \infty ,&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; then &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -19px&amp;quot;&amp;gt;\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.&amp;lt;/math&amp;gt;&lt;br /&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;
|Let &lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{y} &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n.}&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|We then take the natural log of both sides to get&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n\bigg).&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;
|We can interchange limits and continuous functions.&lt;br /&gt;
|-&lt;br /&gt;
|Therefore, we have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\ln y} &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1+\frac{1}{2n}\bigg)^n}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1+\frac{1}{2n}\bigg)}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}.}&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Now, this limit has the form &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -13px&amp;quot;&amp;gt;\frac{0}{0}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Hence, we can use L'Hopital's Rule to calculate this limit.&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;
|Now, we have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\ln y } &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2x}\bigg)}{\frac{1}{x}}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; \overset{L'H}{=} &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{2x}{2x+1}\big(\frac{-1}{2x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{1}{2}\bigg(\frac{2x}{2x+1}\bigg)}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\frac{1}{2}.}&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 4: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|Since &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -5px&amp;quot;&amp;gt;\ln y= 1/2,&amp;lt;/math&amp;gt;&amp;amp;nbsp; we know&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;y=e^{1/2}.&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;
|First, we have&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp;&amp;lt;math&amp;gt;\lim_{n\rightarrow \infty}\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n=\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n.&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, let&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{y} &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n.}&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|We then take the natural log of both sides to get&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\bigg).&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|We can interchange limits and continuous functions.&lt;br /&gt;
|-&lt;br /&gt;
|Therefore, we have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\ln y} &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{1+n}{n}\bigg)^n}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{1+n}{n}\bigg)}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}.}&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Now, this limit has the form &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -13px&amp;quot;&amp;gt;\frac{0}{0}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Hence, we can use L'Hopital's Rule to calculate this limit.&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;
|Now, we have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\ln y } &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{1+x}{x}\bigg)}{\frac{1}{x}}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; \overset{L'H}{=} &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x}{1+x}\big(\frac{-1}{x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{x\rightarrow \infty} \frac{x}{1+x}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{1.}&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 4: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|Since &amp;amp;nbsp;&amp;lt;math style=&amp;quot;vertical-align: -4px&amp;quot;&amp;gt;\ln y= 1,&amp;lt;/math&amp;gt;&amp;amp;nbsp; we know&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;y=e.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Since&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp;&amp;lt;math&amp;gt;\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\neq 0,&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|we have&lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
&amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;amp;nbsp; &amp;lt;math&amp;gt;\begin{array}{rcl}&lt;br /&gt;
\displaystyle{\lim_{n\rightarrow \infty} a_n} &amp;amp; = &amp;amp; \displaystyle{\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\text{DNE}.}&lt;br /&gt;
\end{array}&amp;lt;/math&amp;gt;&lt;br /&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;amp;nbsp; &amp;amp;nbsp;&amp;lt;math&amp;gt;e^{1/2}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|&amp;amp;nbsp; &amp;amp;nbsp; '''(b)''' &amp;amp;nbsp; &amp;amp;nbsp;&amp;lt;math&amp;gt;\text{DNE}&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
[[009C_Sample_Final_3|'''&amp;lt;u&amp;gt;Return to Sample Exam&amp;lt;/u&amp;gt;''']]&lt;/div&gt;</summary>
		<author><name>MathAdmin</name></author>
	</entry>
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