Show that ∑n=1∞(−1)n−1sin(logn)nα\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n-1} \sin(\log n)}{n^{\alpha}}n=1∑∞nα(−1)n−1sin(logn) converges if and only if α>0\alpha > 0α>0.
Hidden so you can work on the problem first.