【答案】B 

【解析】 

\(\left|\sin\frac{n^{3}\pi}{n^{2}+1}\right|=\left|\sin\left(\frac{n^{3}\pi}{n^{2}+1}-n\pi\right)\right|=\left|\sin\frac{n}{n^{2}+1}\pi\right|\sim\frac{n}{n^{2}+1}\pi\sim\frac{1}{n}\pi\) . 

\(\sum_{n = 1}^{\infty}\frac{1}{n}\)发散  \(\therefore\)不是绝对收敛. 

\(\sin\frac{n^{3}\pi}{n^{2}+1}=(-1)^{n}\sin\left(\frac{n^{3}\pi}{n^{2}+1}-n\pi\right)=(-1)^{n}\sin\frac{n}{n^{2}+1}\pi\)，为交错级数. 

\(\sin\frac{n}{n^{2}+1}\pi\)递减，\(\therefore\)条件收敛. 

\(\sum_{n = 1}^{\infty}(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\) . 

\(\left|(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\right|=\left|-\frac{1}{3}\frac{1}{n^{\frac{2}{3}\times\frac{3}{2}}}+o\left(\frac{1}{n^{2}}\right)\right|\)（注：原解析此处\(n\)的次数书写可能有误，推测应为\(\left|-\frac{1}{3}\frac{1}{n^{\frac{4}{3}}}+o\left(\frac{1}{n^{\frac{4}{3}}}\right)\right|\) ，按你提供内容提取 ），实际准确是利用等价无穷小，当\(x\to0\)时，\(x - \tan x\sim-\frac{1}{3}x^{3}\)，令\(x = \frac{1}{\sqrt[3]{n^{2}}}=n^{-\frac{2}{3}}\)，则\(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\sim-\frac{1}{3}n^{-2}\)，所以\(\left|(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\right|\sim\frac{1}{3n^{2}}\)  ） 

\(\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\)收敛 \(\therefore\sum_{n = 1}^{\infty}(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\) 绝对收敛