Answer
Convergent
Work Step by Step
Let $b_n=\frac{n}{2^n}$
For all $n\geq 1$, we have
$n+1\leq 2n$
$\frac{n+1}{2}\leq n$
$\frac{n+1}{2^n\cdot 2}\leq \frac{n}{2^n}$
$\frac{n+1}{2^{n+1}}\leq \frac{n}{2^n}$
and we know that as $n$ is large enough the value of $\frac{n}{2^n}$ is close to zero.
It means $b_{n+1}\leq b_n$ for all $n$ and $\lim\limits_{n \to \infty}b_n=0$.
It follows by the Alternating Series Test with $b_n=\frac{n}{2^n}$ that the series $\sum_{n=1}^\infty (-1)^n\frac{n}{2^n}$ is convergent.
