<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>8.3.16 (Proof that sum 1/n diverges)</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h3>Example</h3>

<p>Show that the series \(\sum\limits_{n=1}^\infty \frac{1}{n}\) diverges.</p>

<h3>Solution</h3>

<p>To prove this fact, notice that the given series is a p-series with \(p=1\). Let us write out the first few terms of the series
\[
S_1= \sum\limits_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots
\] 
We can compare this with the series 
\[
S_2 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \cdots.
\]
Clearly, the second series is smaller than the first. Notice that in \(S_2\) each set of terms wafter the first term can be collected to add up to \(\frac{1}{2}\). Therefore, 
\[
S_2 =  1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \cdots.
\]
Therefore, \(S_2\) diverges. But \(S_2\) is smaller than \(S_1\) and therefore by the comparison test \(S_1 = \sum\limits_{n=1}^\infty \frac{1}{n}\) diverges as well.</p>]]></text>
    </questiontext>
    <generalfeedback format="moodle_auto_format">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
