Convergence of a series involving products of sequences I am trying to determine whether or not the following series converges$$\sum_{n=1}^\infty \frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)}$$My first attempt at figuring this out involved the fact that I know $$\prod_{i=1}^n(2i-1)=\frac{(2n)!}{2^nn!}$$It seems like I could simply change this to $\frac{(2(n+1))!}{2^{n+1}(n+1)!}$ and it would work in this scenario. However, I'm not certain of that, and even if it did work I have no idea how to approach the denominator. What am I missing?
 A: $\sum_\limits{n=1}^\infty \frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)}$
Ratio test:
$\lim_\limits {n\to \infty}\frac {a_{n+1}}{a_n} = \lim_\limits {n\to \infty}\frac {2n+3}{3n+5} = \frac 23
<1$
The series converges.
A: We have:
$$ S=\sum_{n\geq 1}\prod_{k=0}^{n}\frac{2k+1}{3k+2}=\frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}}\sum_{n\geq 1}\frac{\Gamma\left(n+\frac{3}{2}\right)}{\Gamma\left(n+\frac{5}{3}\right)}\left(\frac{2}{3}\right)^n$$
hence the series is convergent by comparison with a convergent geometric series, due to $\Gamma\left(n+\frac{3}{2}\right)<\Gamma\left(n+\frac{5}{3}\right)$. Additionally, by Gautschi's inequality we have
$$ S < \frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}}\sum_{n\geq 1}\frac{\left(\frac{2}{3}\right)^n}{\left(n+\frac{2}{3}\right)^{1/6}} $$
and by exploiting Euler's beta function we have
$$\begin{eqnarray*} S &=& \frac{\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}\,\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2}{3}\right)^n x^{n+\frac{1}{2}}(1-x)^{-\frac{5}{6}}\,dx\\ &=& \color{red}{\frac{2\,\Gamma\left(\frac{5}{3}\right)}{\sqrt{\pi}\,\Gamma\left(\frac{1}{6}\right)}\int_{0}^{1}\frac{x^{3/2}}{(1-x)^{1/6}(3-2x)}\,dx}. \end{eqnarray*}$$
A: Hint:
Since $\frac{2k+1}{3k+2}\le\frac23$
$$
\begin{align}
\frac{1*3*5*...*(2n+1)}{2*5*8*...*(3n+2)}
&\le\left(\frac23\right)^{n+1}
\end{align}
$$
