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.

Convergence for $0<a\leq 1$ is a tricky matter. Since $\sin(\log n)=\Im(n^i)$,
$$\sum_{n=1}^{+\infty}(-1)^{n-1}\frac{\sin(\log n)}{n^a}=\Im\left(\left(1-2^{1+i-a}\right)\zeta(a-i)\right),$$
where $\zeta(s)$ is the Riemann zeta function.
In order to prove the convergence of the LHS, consider that:
$$\left(\frac{\sin(\log n)}{n^a}-\frac{\sin(\log(n+1))}{(n+1)^a}\right)=\frac{\sin\log n-\sin\log(n+1)}{n^a}+\sin(\log(n+1))\cdot\left(\frac{1}{n^a}-\frac{1}{(n+1)^a}\right).$$
Now both terms on the right are $O(n^{-(a+1)})$, since $\frac{d}{dx}\frac{1}{x^a}=\frac{-a}{x^{a+1}}$ and $\frac{d}{dx}\sin\log x=\frac{\cos\log x}{x}$.
