Calculus II - Absolute Convergence

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Section 10.9 : Absolute Convergence

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3. Determine if the following series is absolutely convergent, conditionally convergent or divergent.

$\sum\limits_{n = 3}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)}}{{{n^3} + 1}}}$

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Okay, let’s first see if the series converges or diverges if we put absolute value on the series terms.

$\sum\limits_{n = 3}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right)}}{{{n^3} + 1}}} \right|} = \sum\limits_{n = 3}^\infty {\frac{{n + 1}}{{{n^3} + 1}}}$

We know by the $p$ -series test that the following series converges.

$\sum\limits_{n = 3}^\infty {\frac{1}{{{n^2}}}}$

If we now compute the following limit,

$c = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{n + 1}}{{{n^3} + 1}}\,\frac{{{n^2}}}{1}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^3} + {n^2}}}{{{n^3} + 1}}\,} \right] = 1$

we know by the Limit Comparison Test that the two series in this problem have the same convergence because $c$ is neither zero or infinity and because $\sum\limits_{n = 3}^\infty {\frac{1}{{{n^2}}}}$ converges we know that the series from the problem statement must also converge.

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So, because the series with the absolute value converges we know that the series in the problem statement is absolutely convergent.