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      <text>16.2.1.5 Convergence of the alternating harmonic series</text>
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      <text><![CDATA[<p>As with the harmonic series we shall group the terms of the alternating harmonic series, this time to
display its convergence.&nbsp;</p><p>The alternating harmonic series is:<br></p><p>\[ S =&nbsp;1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dots\]</p><p>This series may be re-grouped in two distinct ways.&nbsp;</p><p><br></p><p><b>1st re-grouping</b></p><p>\[ S = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} + \dots = 1 - \left(\dfrac{1}{2} - \dfrac{1}{3} \right) - \left( \dfrac{1}{4} - \dfrac{1}{5} \right) - \left(&nbsp; \dfrac{1}{6} - \dfrac{1}{7}\right) - \dots\]</p><p>Each term in brackets is positive since \(\dfrac{1}{2}&gt; \dfrac{1}{3}\), \(\dfrac{1}{4}&gt;\dfrac{1}{5}\) and so on. So we easily conclude that \( S&lt; 1\) since we are subtracting only positive numbers from \(1\).&nbsp;</p><p><b><br></b></p><p><b>2nd re-grouping</b></p><p><span>\[ S = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} + \dots = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) +&nbsp;\left(\dfrac{1}{5} - \dfrac{1}{6}\right) + \dots \]<br></span></p><p><span>Again, each term in brackets is positive since \(1 &gt; \dfrac{1}{2}\) , \(\dfrac{1}{3} &gt; \dfrac{1}{4}\) , \(\dfrac{1}{5} &gt; \dfrac{1}{6}\) and so on.<br></span></p><p><span>So we can also argue that \(S &gt; \dfrac{1}{2}\) since we are adding only positive numbers to the value of the first
term, \(\dfrac{1}{2}\) .&nbsp;</span></p><p><span><br></span></p><p><span>The conclusion that is forced upon us is that \[\dfrac{1}{2}&lt;S&lt;1\] so that the&nbsp;alternating series is convergent since its sum, \(S\), lies in the range \(\dfrac{1}{2} \to 1\). It will be shown in a later Section that \(S = \ln{2} = 0.693\).<br></span></p>]]></text>
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