Answer
Converges, by Th. 9.12.
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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Observe that $\displaystyle \frac{1}{\sqrt{n^{3}+1}} < \frac{1}{\sqrt{n^{3}}}=\frac{1}{n^{3/2}}$
$0 < \displaystyle \frac{1}{\sqrt{n^{3}+1}} < \frac{1}{n^{3/2}}$
$(0 < a_{n} \leq b_{n})$
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ converges (p series, $p >1)$
So, by case (1),
$\displaystyle \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{3}+1}}$ also converges.
