Answer
Diverges
Work Step by Step
We are given that $a_k=\dfrac{2^k}{k^{99}}$
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, $r=\lim\limits_{k \to \infty}\dfrac{2^{k+1}k^{99}}{2^k (k+1)^{99}}\\=2 \lim\limits_{k \to \infty} (\dfrac{k}{k+1})^{99} \\=(2) (1)^{99} \\=2$
Therefore, the series diverges by the ratio test.
