Answer
converges
Work Step by Step
Let us consider $u_n=(-1)^{n+1}\dfrac{(0.1)^n}{n}$
and $u_n$ refers to all positive values of $n$.
Also, $|u_n|=\dfrac{(0.1)^n}{n}$
and $\dfrac{(0.1)^n}{n} \lt (0.1)^n$.
We can see that $\Sigma_{n=1}^\infty (0.1)^n$ shows a convergent geometric series, therefore, the series converges absolutely by the direct comparison test.