<s>[INST] Test the series $\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$ Use the ratio test for absolute convergence to determine
whether the series absolutely or
diverge.
$$\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$$
I don't understand how the general term becomes this $$\lim_{k\to\infty} \frac{a_{k+1}}{a_k} = \lim_{k\to\infty} \frac{2k+3}{3k+4}$$
Obviously this Absolutely converges, I just dont get the 2nd step. 
 [/INST] Let us write
$$a_k=\frac{ 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}.$$
Then, we  have
$$\begin{align}
\frac{a_{k+1}}{a_k}&=\frac{1\cdot3\cdot5\cdots [2(k+1)+1]}{1\cdot 4\cdot 7\cdots [3(k+1)+1]}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots (2k+3)}{1\cdot 4\cdot 7\cdots (3k+4)}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots(2k+1) (2k+3)}{1\cdot 4\cdot 7\cdots (3k+1)(3k+4)}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots(2k+1)}{1\cdot 4\cdot 7\cdots (3k+1)}\quad\cdot\quad\frac{2k+3}{3k+4}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{2k+3}{3k+4}.
\end{align}$$
This proves the 2nd step you asked.
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