Answer
Converges
Work Step by Step
We can write the general form of the given series as: $a_k=\dfrac{k^2}{2^k}$
Root Test states that when $\Sigma a_k$ is an infinite series with positive terms and, then $r=\lim\limits_{k \to \infty}\sqrt[k] a_k$
a) When $0 \leq r \lt 1$, the series converges. (b) When $r \gt 1$, or, $\infty$, so the series diverges. (c) When $r=1$, the ratio test is inconclusive.
Now, $r=\lim\limits_{k \to \infty}\sqrt[k] {\dfrac{k^2}{2^k}} \\= \lim\limits_{k \to \infty} \dfrac{(\sqrt[k] k)^2}{2} \\=\dfrac{1}{2}$
Therefore, the series converges by the root test.
