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      <text>16.2.1.4 Divergence of the harmonic series</text>
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      <text><![CDATA[<p>The harmonic series:<br></p><p>\[ 1 + \dfrac{1}{2} +&nbsp;\dfrac{1}{3} +&nbsp;\dfrac{1}{4} + \dfrac{1}{5} + \dots\]&nbsp;</p><p>has a general term \(a_n=\dfrac{1}{n}\) which clearly gets smaller and smaller as \(n \to \infty\). However, surprisingly,
the series is divergent. Its divergence is demonstrated by showing that the harmonic series is greater
than another series which is obviously divergent. We do this by grouping the terms of the harmonic
series in a particular way:&nbsp;</p><p>\[&nbsp;1 + \dfrac{1}{2} +&nbsp;\dfrac{1}{3} +&nbsp;\dfrac{1}{4} + \dfrac{1}{5} + \dots \equiv 1 + \left(\dfrac{1}{2}\right) +&nbsp;&nbsp;\left(\dfrac{1}{3} + \dfrac{1}{4}\right) +&nbsp;&nbsp;\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right) + \dots\]</p><p>Now \[ \begin{aligned}&nbsp;\left(\dfrac{1}{3} + \dfrac{1}{4}\right) &amp; &gt; \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2} \\&nbsp;&nbsp;\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right) &amp; &gt; \dfrac{1}{8} + \dfrac{1}{8} +&nbsp;\dfrac{1}{8} +&nbsp;\dfrac{1}{8} = \dfrac{1}{2} \\ \left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12} + \dfrac{1}{13} + \dfrac{1}{14}+ \dfrac{1}{15}+ \dfrac{1}{16}\right) &amp; &gt;&nbsp;\left(\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16} + \dfrac{1}{16} + \dfrac{1}{16}+ \dfrac{1}{16}+ \dfrac{1}{16}\right) = \dfrac{1}{2} \end{aligned}\]</p><p>and so on. Hence the harmonic series satisfies:</p><p>\[\begin{aligned}1+\dfrac{1}{2} +&nbsp;\dfrac{1}{3} +&nbsp;\dfrac{1}{4} + \dfrac{1}{5} + \dots &amp; \equiv 1 + \left(\dfrac{1}{2}\right) +&nbsp;&nbsp;\left(\dfrac{1}{3} + \dfrac{1}{4}\right) +&nbsp;&nbsp;\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+ \dots \\ &amp; &gt;1 + \dfrac{1}{2} +&nbsp;\dfrac{1}{2} +&nbsp;\dfrac{1}{2} + \dots \end{aligned}\]</p><p>The right hand side of this inequality is clearly divergent so <b>the harmonic series is divergent.</b></p>]]></text>
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