## University Calculus: Early Transcendentals (3rd Edition)

Consider $u_n=(-1)^{n+1}\dfrac{(0.1)^n}{n}$ Here, $u_n$ is positive for all the values of $n$. and $|u_n|=\dfrac{(0.1)^n}{n}\lt (0.1)^n$ We know that $(0.1)^n$ describes a convergent geometric series. Thus the series converges absolutely by the Direct Comparison Test.