Answer
Converges
Work Step by Step
We are given that $a_k=\dfrac{2 \cdot 4^k}{(2k+1)!}$
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, $r=\lim\limits_{k \to \infty} |\dfrac{2^{k+1} k!}{2^k (k+1)!}|\\=\lim\limits_{k \to \infty} \dfrac{\dfrac{8 \cdot 4^k}{(2k+3)(2k+2)(2k+1)!}}{\dfrac{2 \cdot 4^k}{(2k+1)!}}\\= \lim\limits_{k \to \infty} \dfrac{4}{(2k+3)(2k+2)}$
Since, for all $n \gt 0$, we have: $\lim\limits_{x \to \infty} x^n =\infty$ and $\lim\limits_{x \to \infty} x^{-n} =0$
So, $r= \lim\limits_{k \to \infty} \dfrac{4}{(2k+3)(2k+2)}=\dfrac{1}{\infty}=0$
Therefore, the given series converges by the ratio test.
