{\sum_{n=2}^{+\infty} {{\sin(\pi\cdot(n^3+n)^{1\over 3}}) \over \ln^{\alpha}n} =  {\sum_{n=2}^{+\infty}(-1)^{n} {{\sin(\pi\cdot(n^3+n)^{1\over 3}}-\pi n) \over \ln^{\alpha}n}