Answer
Converges
Work Step by Step
Let $u_n=\dfrac{1-n}{n 2^{n}}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{u_{n+1}}{u_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{-n}{(n+1)2^{2n+1}}}{(1-n)/n (2^{n})}|$
$\implies \lim\limits_{n \to \infty} \dfrac{n^2}{2n^2-2}= \lim\limits_{n \to \infty} \dfrac{1}{2-\dfrac{2}{n^2}}=\dfrac{1}{2-0}=\dfrac{1}{2} \lt 1$
Thus, the series converges by the ratio test.