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. 
 A: 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.
A: The given series is absolutely convergent by the ratio test, since by setting $c_k=\frac{(2k+1)!!}{(3k+1)!!!}$ we have $\lim_{k\to\infty}\frac{c_{k+1}}{c_k}=\frac{2}{3}$, hence the radius of convergence of $f(x)=\sum_{n\geq 0}(-1)^n c_n x^n$ is $\frac{3}{2}$. Actually
$$ f(x)=\phantom{}_2 F_1\left(1,\frac{3}{2};\frac{4}{3};-\frac{2x}{3}\right)\approx \frac{28-x}{28+20 x}$$
in a neighbourhood of the origin by Padé approximants, hence the given series is quite close to $\frac{9}{16}$.
