Answer
Converges
Work Step by Step
We are given that $a_k=\dfrac{k^2}{4^k}$
Ratio Test states that when $\Sigma a_k$ is an infinite series with positive terms and, then $r=\lim\limits_{k \to \infty}\dfrac{a_{k+1}}{a_k}$
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, $r=\lim\limits_{k \to \infty}\dfrac{(k+1)^{2} 4^k}{k^2 4^{(k+1)}}\\=\lim\limits_{k \to \infty} \dfrac{(k^2 +2k+1) \cdot 4^k}{k^2 (4)(4^k)}\\=\lim\limits_{k \to \infty} \dfrac{k^2+2k+1}{4k^2}\\=\dfrac{1}{4}$
Therefore, the series converges by the ratio test.
