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.

I have not managed to prove that the sequence
$$
s_n=\sum_{k=1}^n (-1)^{k-1}\sin(\ln k), \quad n\in\mathbb N,
$$
is bounded, which would establish that the series converges for every $\alpha>0$, but it is not hard to see that 
$$
s_n={\mathcal O}(\ln n),
$$
since (due to Mean Value Theorem)
$$
|s_{n+2}-s_n|=\big|\sin(\ln (n+2))-\sin(\ln(n+1))\big| \le |\ln(n+2)-\ln n|=\ln\left(1+\frac{2}{n}\right)\le \frac{2}{n},
$$
and thus
$$
|s_n| \le 2\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)\le 2\ln (n+1).
$$
Therefore
\begin{align}
\sum_{j=1}^n \frac{(-1)^{j-1}}{j^\alpha}\sin(\ln j)&=\sum_{j=1}^n \frac{s_j-s_{j-1}}{j^\alpha}=
\sum_{j=1}^n\frac{s_{j}}{j^\alpha}-\sum_{j=0}^{n-1}\frac{s_{j}}{(j+1)^\alpha} \\ &=\sum_{j=1}^{n-1}s_j\left(\frac{1}{j^\alpha}-\frac{1}{(j+1)^\alpha}\right)+\frac{s_n}{n^\alpha}.
\end{align}
The second term $\frac{s_n}{n^\alpha}={\mathcal O}(n^{-\alpha}\ln n)$ tends to zero, while the first term is a convergent series, due to the comparison test. Indeed
$$
\left|s_j\left(\frac{1}{j^\alpha}-\frac{1}{(j+1)^\alpha}\right)\right|\le \frac{2\alpha\ln j}{j^{\alpha+1}},
$$ 
and
$$
\sum_{j=1}^\infty \frac{2\alpha\ln j}{j^{\alpha+1}} <\infty.
$$
