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      <text>16.4.2 The radius of convergence</text>
    </name>
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      <text><![CDATA[<h4>Radius of convergence</h4>
<p>The most important statement one can make about a power series is that there exists a number, \(R\), called the radius of convergence, such that if \(|x| &lt; R\) the power series is absolutely convergent and if \(|x| &gt; R\) the power series is divergent.
    At the two points \(x = −R\) and \(x = R\) the power series may be convergent or divergent.
</p>

<p></p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p>For a power series \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) with radius of convergence \(R\) then </p>
    <li> the series converges absolutely if \(|x|&lt;R\)</li>
    <li> the series diverges if \(|x|&gt;R\)
    </li>
    <li> the series may be convergent or divergent at \(x=\pm R\)</li>
    <figure class="HELM_figure"><img src="@@PLUGINFILE@@/16-4-KeyPoint11.svg" alt="Convergence of power series" width="455" height="65" class="img-responsive atto_image_button_text-bottom"></figure>
</div>

<p>For any particular power series \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) the value of \(R\) can be obtained using the ratio test.&nbsp;</p>
<p>To do this, we apply the ratio test for a general \(x\), we can think of this as fixing \(x\) and asking if the series is convergent. We know, from the ratio test that \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) is absolutely convergent if \[ \lim_{p
    \to \infty} \dfrac{|b_{p+1} x^{p+1}|}{|b_p x^p|} = \lim_{p \to \infty}\left|\dfrac{b_{p+1}}{b_p}\right| |x|&lt;1 \quad \mathrm{implying} \quad |x|&lt;\lim_{p\to\infty}\left|\dfrac{b_p}{b_{p+1}}\right| \qquad \mathrm{and \; so} \quad R = \lim_{p\to\infty}\left|\dfrac{b_p}{b_{p+1}}\right|&nbsp;\]
</p>
<p>Warning: Note that the terms appearing in the numerator and denominator for the ratio are reversed in the definition of \(R\). This for a good reason, it happens when we divide in the final step above.</p>
<p>It can help to remember that the ratio test looks at how the terms grow by considering the ratio of a term with the previous term. If this is less than \(1\) then the terms are decreasing which is necessary for convergence. We essentially compare the
    series with a geometric series. In a power series, it looks a little bit like a geometric series with common ratio \(x\), the series converges if the growth rate \(x\) is sufficiently slowed by the growth (or decay) of the coefficients. Hence the
    series converges if the growth of \(x\) is less than "one over the growth of the coefficients" as the product needs to be less than \(1\).</p>
<p>If nothing else, take care to return to the definitions when you need.</p>

<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts">
    <li>Find the radius of convergence of the series \[1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots\] </li>
    <li>Investigate what happens at the end-points \(x=-1, x=+1\) of the region of absolute convergence.</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>Here \[1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots = \sum_{p=0}^\infty \dfrac{x^p}{p+1} \] so \[b_p = \dfrac{1}{p+1} \quad \therefore \quad b_{p+1}=\dfrac{1}{p+2}.\] In this case, \[R= \lim_{p \to \infty}\left| \dfrac{p+2}{p+1}\right|
        = 1\] so the given series is absolutely convergent if \(|x|&lt;1\) and is divergent if \(|x|&gt;1\). </li>
    <li>
        <p>At \(x=+1\) the series is \[1+\dfrac{1}{2} + \dfrac{1}{3} + \cdots\] which is divergent (the harmonic series). However, at \(x=-1\) the series is \[1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots \] which is convergent (the alternative harmonic
            series).
        </p>
        <p> Finally, therefore, the series \[ 1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots \] is convergent if \(-1\leq x &lt; 1\). </p>
    </li>
</ol>]]></text>
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<file name="Screenshot 2020-10-20 at 12.06.43.png" path="/" encoding="base64">[encoded data, 29236 characters]</file>
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