$\sum_{k=1}^\infty \frac {(-1)^{k+1}\cdot 1\cdot 4\cdot 7\cdots(3k-2)}{2^k\cdot k!}$

I'm working on:

$$\sum_{k=1}^\infty \frac {(-1)^{k+1}\cdot 1\cdot 4\cdot 7\cdots(3k-2)}{2^k\cdot k!}$$

I've already shown that this series doesn't absolutely converge. I can't use Abel's test because though the sequence $b_k = \frac {1}{2^k\cdot k!}$ is monotone and bounded, $$\sum_{k=1}^\infty (-1)^{k+1}\cdot 1\cdot 4\cdot 7\cdots(3k-2)$$ doesn't converge.

Taking the limit of the sequence as $k \to \infty$ looks daunting without some means of specifying the numerator in a nicer form. I can't do one of the comparison tests because the terms aren't all positive. I'm at a loss as to how to show that this series either diverges or converges conditionally.

• $\displaystyle\sum_{k=0}^\infty\frac{1\cdot4\cdot7\cdots(3k-2)}{k!}~x^k ~=~ \frac1{\sqrt[\large^3]{1-3x}}~:~$ see binomial series. In this case, $x=-\dfrac12~.~$ However, the series only converges for $x\in\bigg[-\dfrac13~,~\dfrac13\bigg).~$ – Lucian Apr 25 '15 at 10:39

What is the behavior of $|a_{k+1}/a_k|$ as $k\rightarrow\infty$?