Let $f(x)=e^x-1-x$, then $f'(x)=e^x-1$ and the only $x\in\mathbb{R}$ so that $e^x-1=0$ is $x=0$. Note that $f''(0)=1$ so $f(x)$ reaches a minimum of $0$ at $x=0$. Therefore, for all $x\in\mathbb{R}$,
$$
1+x\le e^x\tag{1}
$$
Thus, we get that for all $x\in\mathbb{R}$, $x<e^x$, and therefore, for all $x>0$, $$
\log(x)< x\tag{2}
$$
For any $\alpha>0$, apply inequality $(2)$ to $x^\alpha$ and divide by $\alpha$ to get
$$
\log(x)< \tfrac1\alpha x^\alpha\tag{3}
$$
Then for any $\epsilon>0$, we have
$$
\begin{align}
\sum_{k=1}^\infty\frac{\log(k)}{k^{1+\epsilon}}
&<\sum_{k=1}^\infty\frac{\frac2\epsilon k^{\epsilon/2}}{k^{1+\epsilon}}\\
&=\frac2\epsilon\sum_{k=1}^\infty\frac1{k^{1+\epsilon/2}}\tag{4}
\end{align}
$$
and $(4)$ converges for all $\epsilon>0$. Thus, for any $\alpha>1$,
$$
\sum_{k=1}^\infty\frac{\log(k)}{k^\alpha}\tag{5}
$$
converges.
Alternatively, since $\frac{\log(x)}{x^\alpha}$ is a decreasing function when $x\ge e$ and $\alpha\ge1$,
$$
\begin{align}
\sum_{k=3}^\infty\frac{\log(k)}{k^\alpha}
&\le\int_1^\infty\frac{\log(x)}{x^\alpha}\mathrm{d}x\\
&=\int_0^\infty\frac{x}{e^{\alpha x}}\,e^x\,\mathrm{d}x\\
&=\frac1{(\alpha-1)^2}\int_0^\infty x\,e^{-x}\,\mathrm{d}x\\
&=\frac1{(\alpha-1)^2}\tag{6}
\end{align}
$$
Thus, for any $\alpha>1$, we again have that $(5)$ converges.
