calculus - Test the series $sum_{k=0}^{infty}frac{(-1)^k 1 cdot3cdot5 cdots(2k+1)}{1cdot4cdot7cdots(3k+1)}$

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.

Answer

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.