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.
 A: @nanchangjian: 
You are not understanding the cases $a>1$ and $a\leq 0$  correctly.
The reason why it converges for $a>1$: is as @DonAntonio pointed out, 
$$\left|(-1)^{n-1}\frac{\sin\log n}{n^a}\right|\le\frac1{n^a}$$
Notice that $(-1)^{n-1}$ is inside the absolute value sign. 
In case $a\leq 0$: it diverges because 
$$\limsup \left((-1)^{n-1}\frac{\sin\log n}{n^a}\right)\geq 1$$
The case when $0<a\leq 1$: 
Notice that the series is the imaginary part of the following:
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{a-i}}$$
where $i=\sqrt{-1}$. 
So, we consider the Dirichlet series 
$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^s}$$
This series has $1$ as abscissa of absolute convergence and $0$ as abscissa of convergence. 
That means the series converges whenever $\textrm{Re} (s) >0$. 
Since $\textrm{Re}(a-i) = a >0$, the series converges. 
It will be helpful if you understand this lemma about Dirichlet series:
(Lemma)
Suppose that $F(s)=\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges for some $s_0\in\mathbb{R}$. Then the series converges for all $s$ in the region $\textrm{Re}(s)>s_0$. 
The proof of this uses partial summation. 
A: 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.
A: 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 $\frac{d}{dx}\sin\log x=\frac{\cos\log x}{x}$.
A: 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
$$
\sum_{j=1}^\infty \frac{2\alpha\ln j}{j^{\alpha+1}} <\infty.
$$
A: DonAntonio's hint along with the Alternating Series Test (http://en.wikipedia.org/wiki/Alternating_series_test) should give you all the values of $a$ for which your series converges.
A: Conjecture. The series converges if and only if $\alpha>0$. 
It suffices to show that the sequence $s_n=\sum_{k=1}^n (-1)^{k-1}\sin(\ln k)$, $n\in\mathbb N,$
is bounded. For if $|s_n|\le M$, for some $M>0$, then
\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}
which clearly converges.
Using quadruple precision (real*16) FORTRAN code, for $n=1,\ldots,10^8$, I got the bounds $s_n\in(-.271,.732)$. It remains to show analytically that $s_n$ is bounded. 
A: @Yiorgos S. Smyrlis
Your conjecture is true.
A sketch:
$\frac{\sin(\ln(n))}{n^a} -\frac{\sin(\ln(n-1))}{(n-1)^a} =^{\text{ By MVT}} \frac{ \cos(\ln(t))-a \sin(\ln(t))}{t^{a+1}} $ for some $t \in ]n,n-1[$.
 Now all we need to do is rewrite the series as 
$ \lim_{t \to \infty}\sum_{n=1}^{t} \left( \frac{\sin(\ln(2n+1))}{(2n+1)^a} -\frac{\sin(\ln(2n))}{(2n)^a}\right)$ 
From here, the series clearly converge for all $a>0$
