Does the series $\sum_{n=1}^\infty (-1)^n \frac{\cos(\ln(n))}{n^{\epsilon}},\,\epsilon>0$ converge? Numerical results showed that the series
$$Q=\sum_{n=1}^{\infty} (-1)^n \frac{\cos(\ln(n))}{n^{\epsilon}},\epsilon>0\tag{1}$$
with $\epsilon=10^{-6}$ converged to $-0.53259554096828...$
We can rewrite this series as:
$$Q=\sum_{k=0}^{\infty} a_k\tag{2}$$
$$a_k=\frac{\cos(\ln(2k+1))}{(2k+1)^{\epsilon}}-\frac{\cos(\ln(2k+2))}{(2k+2)^{\epsilon}}\tag{3}$$
Does this series converge?
Using a method in a related question, we can show that (with $m=2k+1$)
$$-a_k=m^{-1-\epsilon}\left(\epsilon\cos(\ln m)+\sin(\ln m)\right)+O(m^{-2-\epsilon}).\qquad m\to\infty \tag{4}$$
Thus
$$|a_k|\le m^{-1-\epsilon}\left(\epsilon|\cos(\ln m)|+|\sin(\ln m)|\right)+O(m^{-2-\epsilon})=O(m^{-1-\epsilon})=O(k^{-1-\epsilon})\tag{5}$$
Therefore the series in (1) is convergent.
Am I right?
Thanks-
mike
 A: Let 
$$
\eta(s)=\sum_{n=1}^\infty (-1)^{n+1}n^{-s},
$$
with $s=\sigma+it$.  The real part of this series is
$$
-\sum_{n=1}^\infty (-1)^n n^{-\sigma} \cos(t\ln(n)).
$$
With $t=0$, the series converges (conditionally) for $\sigma>0$ by the alternating series test.  A standard theorem about Dirichlet series then says the series converges (conditionally) for complex $s$ with $\text{Re}(s)>0$.  If I recall correctly, the convergence is uniform in sectors $|\arg(s)|<\pi/2-\epsilon$.
It's worth noting that $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$.
A: Let me provide a more elementary answer. 
As observed in the OP, we can rewrite the series as
$$
Q=\sum_{k=0}^{\infty} a_k,
$$
where
$$
a_k=\frac{\cos(\ln(2k+1))}{(2k+1)^{\epsilon}}-\frac{\cos(\ln(2k+2))}{(2k+2)^{\epsilon}}.
$$
Now
$$
a_k=\frac{(2k+2)^{\epsilon}\cos(\ln(2k+1))-(2k+1)^{\epsilon}\cos(\ln(2k+2))}{\big((2k+1)(2k+2)\big)^{\epsilon}}\\=\frac{(2k+2)^{\epsilon}\big(\cos(\ln(2k+1))-\cos(\ln(2k+2))\big)}{\big((2k+1)(2k+2)\big)^{\epsilon}}
+
\frac{\big((2k+2)^{\epsilon}-(2k+1)^{\epsilon}\big)\cos(\ln(2k+2))}{\big((2k+1)(2k+2)\big)^{\epsilon}} \\ = b_k+c_k.
$$
Now Mean Value Theorem for $f(x)=\cos(\ln x)$, with $f'(x)=-\sin(\ln x)/x$, provides that
$$
\cos(\ln(2k+1))-\cos(\ln(2k+2))=\frac{\sin(\ln (2k+1+\xi_k))}{2k+1+\xi_k},
$$
for some $\xi_k\in(0,1)$, and thus
$$
\lvert b_k\rvert\color{red}{=}
\frac{\lvert\sin(\ln (2k+1+\xi_k))\rvert}{(2k+1+\xi_k)(2k+1)^\epsilon}\le
\frac{1}{(2k+1)^{1+\epsilon}},
$$
which implies that $\sum_{k=0}^\infty \lvert b_k\rvert<\infty$. Meanwhile, applying the Mean Value Theorem once again to $g(x)=x^\epsilon$, we obtain that
$$
\lvert c_k\rvert \le \frac{(2k+2)^{\epsilon}-(2k+1)^{\epsilon}}{\big((2k+1)(2k+2)\big)^{\epsilon}}\color{red}{=}\frac{\epsilon(2k+1+\xi_k)^{\epsilon-1}}{\big((2k+1)(2k+2)\big)^{\epsilon}}, \quad \xi_k\in(0,1),
$$
and hence
$$
\lvert c_k\rvert \le \frac{\epsilon(2k+2)^{\epsilon-1}}{\big((2k+1)(2k+2)\big)^{\epsilon}}
=\frac{\epsilon}{(2k+1)^{\epsilon}(2k+2)}\le\frac{\epsilon}{(2k+1)^{1+\epsilon}},
$$
which also implies that $\sum_{k=0}^\infty \lvert c_k\rvert<\infty$.
