The task is to determine for which values of $\alpha$ is the following series is conditionally convergent and absolutely convergent. My attempt is below.
$$\sum_{n=1}^{\infty} {n^{-\alpha}\cdot(\ln{n}) \cdot (\sin{n})}$$
I first noted that if $\alpha \leq 0$ then the limit is not $0$ thus we must have that $\alpha \gt 0$.
Next, I decided to check for absolute convergence. Using Dirichlet's test, it is known that the partial sums of the sine function are bounded (see here for an example of a proof). Thus, what remains to show is that $\dfrac {\ln{n}} {n^\alpha}$ is monotone decreasing since the limit is clearly $0$ for $\alpha\gt 0$.
I calculated the derivative:
$$f(n)=\dfrac {\ln{n}} {n^\alpha} \\ \therefore f'(n) = \dfrac {n^{\alpha -1}\left(1-\alpha\ln{n}\right) } {n^{2\alpha}} \tag{$*$}$$
Now,
$$f'(n) \lt 0\iff n^{\alpha -1}\left(1-\alpha\ln{n}\right) < 0$$ since $\displaystyle{ \forall \alpha, \qquad n^{2\alpha} > 0 }$.
From $(*)$ we can see that as $n \to \infty$, the expression $\left(1-\alpha\ln{n}\right)$ will eventually become negative for all $\alpha > 0$. Thus the series is absolutely convergent for all $\alpha > 0$. I'd appreciate any feedback on whether or not this solution is valid, and if I can conclude from this that there does not exist an $\alpha$ for which the series is conditionally convergent. I have tried to think of cases I might be missing but so far have found none.
With the help of Mr Gabriel Romon I was able to fill in the missing parts of the solution.
This continues from the proof in the original post, which proves that for $\alpha > 0$, the series is convergent and divergent otherwise.
Let us now check for absolute convergence:
For $\alpha > 1 $ consider $$\sum_{n=1}^{\infty} |\sin(n)| \cdot \left| \frac{\ln n}{n^\alpha} \right| \leq \sum_{n=1}^{\infty} 1\cdot \left| \frac{\ln n}{n^\alpha} \right| $$
Now, since $\alpha > 1 $ and $\displaystyle \frac{\ln n}{n^{(\alpha-1)/2}} \to 0 $ as $n \to \infty$ we have that
$$\dfrac{\ln n}{n^\alpha} =\dfrac{\ln n}{n^{(\alpha-1)/2}} \dfrac{1}{n^{(\alpha+1)/2}} \leq M \cdot \dfrac{1}{n^{(\alpha+1)/2}} $$
Therefore
$$\sum_{n=1}^{\infty} |\sin(n)| \cdot \left| \frac{\ln n}{n^\alpha} \right| \leq \sum_{n=1}^{\infty} 1\cdot \left| \frac{\ln n}{n^\alpha} \right| \leq M \sum_{n=1}^{\infty} \dfrac{1}{n^{(\alpha+1)/2}}$$
The sum $\sum_{n=1}^{\infty} \dfrac{1}{n^{(\alpha+1)/2}}$ is convergent since $\alpha >1 $ therefore the initial sum is absolutely convergent for $\alpha > 1$.
For $\alpha \in (0;1]$ we can use the inequality $$|\sin n|\ge\sin^2n=\frac12\left(1-\cos(2\,n)\right)$$
to get
$$ \sum_{n=1}^{\infty} |\sin(n)| \cdot \left| \frac{\ln n}{n^\alpha} \right| \geq \frac 12 \left(\sum_{n=1}^{\infty} \left| \frac{\ln n}{n^\alpha} \right| - \sum_{n=1}^{\infty} \frac{\cos(2n)}{n^{\alpha}}\right) $$
We can easily verify using Dirichlet's test that $$\sum_{n=1}^{\infty} \dfrac{\cos(2n)}{n^{\alpha}}$$ is convergent for $\alpha \in (0;1]$.
As for $$ \sum_{n=1}^{\infty} \left| \frac{\ln n}{n^\alpha} \right| \tag{$*$}$$ we can note that
$$ \sum_{n>e} \dfrac{\ln n}{n^\alpha} \geq \sum_{n>e} \dfrac{1}{n^\alpha} $$
and
$$\sum_{n>e} \dfrac{1}{n^\alpha} $$ is divergent for $\alpha \in (0;1]$
Therefore, we can conclude that the initial series is conditionally convergent for $\alpha \in (0;1]$.