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Section 10.10 : Ratio Test

1. Determine if the following series converges or diverges.

\[\sum\limits_{n = 1}^\infty {\frac{{{3^{1 - 2n}}}}{{{n^2} + 1}}} \]

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We’ll need to compute \(L\).

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{3^{1 - 2\left( {n + 1} \right)}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^{1 - 2n}}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{3^{ - 1 - 2n}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^{1 - 2n}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{1}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^2}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^2} + 1}}{{9\left[ {{{\left( {n + 1} \right)}^2} + 1} \right]}}} \right| = \frac{1}{9}\end{align*}\]

When computing \({a_{n + 1}}\) be careful to pay attention to any coefficients of \(n\) and powers of \(n\). Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer!

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Okay, we can see that \(L = \frac{1}{9} < 1\) and so by the Ratio Test the series converges.