Use the Direct Comparison Test or Limit Comparison Test to determine the convergence or divergence of the series sum_{n = 1}^{infty}(n/sqrt{n^{}3 + 3n}).

We will use the Limit Comparison Test to determine the convergence or divergence of the series. Let {eq}a_n=frac{n}{sqrt{n^3+3n}}. {/eq} Then, we can simplify {eq}a_n {/eq} as follows: {eq}begin{align*} a_n&=frac{n}{sqrt{n^3+3n}}\ &=frac{n}{sqrt{n(n^2+3)}}\ &=frac{1}{sqrt{n^2+3}}. end{align*} {/eq} Since {eq}sqrt{n^2+3}gt n {/eq} for all {eq}ngeq 1,{/eq} we have {eq}0lt a_n lt \frac{1}{n}. {/eq} Thus, we can use the Limit Comparison Test with the series {eq}sum_{n=1}^{infty}frac{1}{n}. {/eq} Since {eq}sum_{n=1}^{infty}frac{1}{n} {/eq} is a $p-$series with {eq}p=1,{/eq} it is known to diverge. Therefore, by the Limit Comparison Test, we have that the series {eq}sum_{n=1}^{infty}frac{1}{sqrt{n^2+3}} {/eq} also diverges.