You can use the integral test. For
$$
\int \ln(1+x^{-k})\, dx,
$$
use integration by parts with $dv=1\,dx$. This gives
$$
\int \underbrace{\ln(1+x^{-k})}_u\underbrace{1\cdot dx\vphantom{(}}_{dv} =
\underbrace{\ln(1+x^{-k})}_u \cdot \underbrace{x\vphantom{(}}_v- \int
\underbrace{\vphantom{k\over x}x}_v\cdot
\underbrace{{-kx^{-k-1}\over 1+x^{-k}}\,dx }_{du} .
$$
A bit of simplification gives:
$$
\int \ln(1+x^{-k})\, dx=\underbrace{\vphantom{\int} x\ln(1+x^{-k})}_{=A} +\underbrace{\int { k \over x^k+1}\,dx}_{=B}
$$
Note that $A$ and the integrand in $B$ are positive for positive $x$.
The integrand in $B$ satisfies $$ {k\over 2x^k}<{k\over 1+x^k}\le {k\over x^k}.$$
Thus $\int_1^\infty {k\over x^k+1}\,dx$ converges if and only if $k>1$. This implies the series diverges for $k\le 1$.
For $k>1$,
$$\eqalign{\lim_{x\rightarrow\infty} A
&=\lim_{x\rightarrow\infty} {\ln(1+x^{-k})\over 1/x}\cr
&=\lim_{x\rightarrow\infty} {-kx^{-k-1}/(1+x^{-k}) \over -1/x^2}\cr
&= \lim_{x\rightarrow\infty} {kx^{-k+1}\over1+x^{-k}}\cr
&= 0.
}$$
This, together with the previous observation that $\int_1^\infty {k\over x^k+1}\,dx$ converges for $k>1$ implies that the series converges for $k>1$.