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	<title>009C Sample Final 1, Problem 3 - Revision history</title>
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		<title>MathAdmin: Created page with &quot;&lt;span class=&quot;exam&quot;&gt;Determine whether the following series converges or diverges.  ::::::&lt;math&gt;\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}&lt;/math&gt;  {| class=&quot;mw-collapsible mw-col...&quot;</title>
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		<updated>2016-04-19T01:18:30Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;lt;span class=&amp;quot;exam&amp;quot;&amp;gt;Determine whether the following series converges or diverges.  ::::::&amp;lt;math&amp;gt;\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}&amp;lt;/math&amp;gt;  {| class=&amp;quot;mw-collapsible mw-col...&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;Determine whether the following series converges or diverges.&lt;br /&gt;
&lt;br /&gt;
::::::&amp;lt;math&amp;gt;\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^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;
|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.''' If a series absolutely converges, then it also converges. &lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
'''Solution:'''&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;
|We proceed using the ratio test. &lt;br /&gt;
|-&lt;br /&gt;
|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{(-1)^{n+1}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{(-1)^n 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!}{n!}\frac{n^n}{(n+1)^{n+1}}\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^n}{(n+1)(n+1)^n}\bigg|}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}\bigg|\bigg(\frac{n}{n+1}\bigg)^n\bigg|}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^n.}\\&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;
|Now, we continue to calculate the limit from Step 1. 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}{n+1}\bigg)^n}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}e^{\ln(\frac{n}{n+1})^n}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}e^{n\ln(\frac{n}{n+1})}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{e^{\lim_{n \rightarrow \infty}n\ln(\frac{n}{n+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 3: &amp;amp;nbsp;&lt;br /&gt;
|-&lt;br /&gt;
|Now, we need to calculate &amp;lt;math&amp;gt;\lim_{n \rightarrow \infty}n\ln\bigg(\frac{n}{n+1}\bigg).&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|First, we write the limit as &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::&amp;lt;math style=&amp;quot;vertical-align: -16px&amp;quot;&amp;gt;\lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Now, we use L'Hopital's Rule to get &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}n\ln\bigg(\frac{n}{n+1}\bigg)} &amp;amp; \overset{l'H}{=} &amp;amp; \displaystyle{\lim_{n \rightarrow \infty}\frac{\frac{n+1}{n}\frac{(n+1)-n}{(n+1)^2}}{-\frac{1}{n^2}}}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n(n+1)}(-n^2)}\\&lt;br /&gt;
&amp;amp;&amp;amp;\\&lt;br /&gt;
&amp;amp; = &amp;amp; \displaystyle{\lim_{n \rightarrow \infty} \frac{-n}{n+1}}\\&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;
|We go back to Step 2 and use the limit we calculated in Step 3. &lt;br /&gt;
|-&lt;br /&gt;
|So, we have &lt;br /&gt;
|-&lt;br /&gt;
|&lt;br /&gt;
::&amp;lt;math&amp;gt;\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}&amp;lt;1.&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
|Thus, the series absolutely converges by the Ratio Test.&lt;br /&gt;
|-&lt;br /&gt;
|Since the series absolutely converges, the series also converges.&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; The series converges.&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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