Answer
By the root test, the series absolutely converges.
Work Step by Step
$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{b^{n}_{n}cos(n\pi)}{n}$
Replace $cosn \pi$ with $(-1)^{n}$
$\sum_{n=1}^{\infty}\frac{b^{n}_{n}cos(n\pi)}{n}=\sum_{n=1}^{\infty}\frac{b^{n}_{n}(-1)^{n}}{n}$
$|a_{n}|=|\frac{b^{n}_{n}(-1)^{n}}{n}|=\frac{b^{n}_{n}}{n}$
$\lim\limits_{n \to \infty} \sqrt[n] {|a_{n}|}=\lim\limits_{n \to \infty} \sqrt[n] {\frac{b^{n}_{n}}{n}}$
$=\lim\limits_{n \to \infty} {\frac{b_{n}}{n^{1/n}}}$
$= {\dfrac{\lim\limits_{n \to \infty}b_{n}}{\lim\limits_{n \to \infty}n^{1/n}}}$
$=\frac{1/2}{1}$
$=\frac{1}{2}\lt 1$
By the root test, the series absolutely converges.