How to find whether this series converges or diverges?

Question

Let's suppose I have been given a series that looks like this:

$$\sum_{n=1}^n\frac{1\cdot 3\cdot 5\cdot\cdots\cdot(2n-1)}{2\cdot5\cdot8\cdot\cdots\cdot(3n-1)}$$

What I have been thinking of doing this whole time is breaking down this problem into three sub-problems: generating a formula for each of the numerator and denominator, then combining them into one formula and applying the ratio test on it. So far I have solved for the formula of the numerator, which is:

$$\frac{(2n)!}{(2^n)n!}$$

However the denominator is a bit more tricky. Am I on the right track or is there a better approach to figuring out whether this formula is convergent or divergent?

Answer 1

If you want to verify the convergence of the series whose $n$th term is given by this formula, you can apply the ratio test directly without converting it to a more concise formula. You will find $$\frac{a_{n+1}}{a_n}=\frac{2(n+1)-1}{3(n+1)-1}=\frac{2n+1}{3n+2}=\frac{2+1/n}{3+2/n}\to\frac{2}{3}.$$

Answer 2

It is also interesting to evaluate the infinite series. Since:

$$ \prod_{k=1}^{n}\frac{2k-1}{3k-1}=\frac{2^n\,\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\frac{2}{3}\right)}{3^n\,\Gamma\left(n+\frac{2}{3}\right)\Gamma\left(\frac{1}{2}\right)}\tag{1}$$ we have: $$\begin{eqnarray*} \sum_{n\geq 1}\prod_{k=1}^{n}\frac{2k-1}{3k-1}&=&\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n\,B\left(n+\frac{1}{2},\frac{1}{6}\right)\\&=&\frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\frac{2\sqrt{u}}{3-2u}(1-u)^{-5/6}\,du\\&=&\frac{12\cdot\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\frac{\sqrt{1-x^6}}{1+2x^6}\,dx\tag{2}\end{eqnarray*} $$ with many identities about the hypergeometric $\phantom{}_2 F_1$ function.