# Is the series #\sum_(n=1)^\inftyn^2/(n^3+1)# absolutely convergent, conditionally convergent or divergent?

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(Use the appropriate test)

(Use the appropriate test)

To determine whether the series ( \sum_{n=1}^\infty \frac{n^2}{n^3+1} ) is absolutely convergent, conditionally convergent, or divergent, we need to analyze its convergence behavior.

First, consider the series ( \sum_{n=1}^\infty \frac{n^2}{n^3+1} ). As ( n ) approaches infinity, the terms of the series behave like ( \frac{n^2}{n^3} = \frac{1}{n} ).

Now, the series ( \sum_{n=1}^\infty \frac{1}{n} ) is a p-series with ( p = 1 ). It is a known result that this series diverges.

Since ( \frac{n^2}{n^3+1} ) is bounded by ( \frac{1}{n} ), and ( \sum_{n=1}^\infty \frac{1}{n} ) diverges, then by the Comparison Test, the series ( \sum_{n=1}^\infty \frac{n^2}{n^3+1} ) also diverges.

Thus, the series ( \sum_{n=1}^\infty \frac{n^2}{n^3+1} ) is divergent.

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Diverges by the Limit Comparison Test

For the comparison sequence, we need one whose series' convergent or divergent behavior we know. So, we'll say

So,

Thus, as both series must diverge because of this result, the series diverges by the Limit Comparison Test.

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