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      <text>8.3.17 (Divergence test)</text>
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      <text><![CDATA[<p>Let 
\[
a_n = \frac{{@aa1@}n^2 + {@aa2@}n}{{@bb1@}n^2-{@bb2@}}.
\] 
Is the following statement true?</p>

<p>The series \(\sum\limits_{n=1}^\infty a_n\) converges by the divergence test. [[input:ans1]] [[validation:ans1]]</p>]]></text>
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      <text><![CDATA[<p>Recall that the divergence test tells us that if \(\lim\limits_{n\to\infty} a_n \not=0\) then \(\sum\limits_{n=1}^\infty a_n\) diverges. So calulating this limit here we have 
\[
\frac{{@aa1@}n^2 + {@aa2@}n}{{@bb1@}n^2-{@bb2@}} = \frac{{@aa1@} + \frac{{@aa2@}}{n}}{{@bb1@}-\frac{{@bb2@}}{n^2}}
\]
Thus \(\lim\limits_{n\to\infty}a_n = \frac{@aa1@}{@bb1@} \not= 0\)
Therefore this series diverges.</p>]]></text>
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      <text><![CDATA[aa1: 1+rand(10);
aa2: 1+rand(10);
bb1: 1+rand(10);
bb2: 1+rand(10);

tac1:[[A, false, "A. true"],  [B, true, "B. false"], [C, false, "C. we cannot tell"]];
taa1:B;]]></text>
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      <text>[[feedback:prt1]]</text>
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    <questionnote format="moodle_auto_format">
      <text>\(a_n = \frac{{@aa1@}n^2 + {@aa2@}n}{{@bb1@}n^2-{@bb2@}}\) diverges</text>
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