Study convergence of the series of the $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin{(\ln{n})}}{n^a}$ Question:
Study convergence of the series 
$$\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\sin{(\ln{n})}}{n^a},a\in R$$
My try: Thank you  DonAntonio help, 
(1):$a>1$,
since 
$$|\sin{(\ln{n})}|\le1$$
so
$$\left|\dfrac{\sin{(\ln{n})}}{n^a}\right|\le \dfrac{1}{n^a}$$
Use  Alternating Series Test
we have  $$\sum_{n=1}^{\infty}(-1)^n\dfrac{1}{n^a},a>1$$ is converge.
(2): $a\le 0$
then we have
$$\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{\sin{(\ln{n})}}{n^a}\to \infty$$
because
$$|(-1)^n\sin{(\ln{n})}|\le 1,\dfrac{1}{n^a}\to \infty,n\to\infty$$
for this case
(3) $0\le a\le 1$  
we can't solve it,
Thank you for you help.

Use absolute value
$$\left|(-1)^{n-1}\frac{\sin\log n}{n^a}\right|\le\frac1{n^a}$$
and thus your series converges absolutely for $\;a>1\;$ .
It's clear that for $\;a\le 0\;$ the series diverges (why?), so we're left only with the case $\;0<a\le1\;$...and perhaps later I'll think of something.
