Prove that the series is conditionally convergent $$ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $$
I have done the$ \sum_{n=1}^\infty  \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ part , and showed it divergent using Gauss test .
But i am not able to do this part $ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $  ,tried leibniz test to do , but could not do that.
I have no idea how to do this please help.
 A: $(\frac{1}{2})^2 < \frac{1}{2}$,
$(\frac{4}{5})^2 < \frac{3}{4}$,
$(\frac{7}{8})^2 < \frac{5}{6}$, $\dots \dots$
$(\frac{3n-2}{3n-1})^2 < \frac{2n-1}{2n}$
Multiplying all , we get 
$\prod_{n=1}^n(\frac{3n-2}{3n-1})^2 <\prod_{n=1}^n \frac{2n-1}{2n}$$  ......(1)$
Now, 
$(\frac{1}{2}) < \frac{2}{3}$,$(\frac{3}{4}) < \frac{4}{5}$,$\dots$ $(\frac{2n-1}{2n}) < \frac{2n}{2n+1}$
Multiplying all, we get
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n}) < \frac{2.4.6. \dots2n}{3.5.\dots.2n-1.2n+1}$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n}) < \frac{2.4.6. \dots2n}{3.5.\dots.2n-1}.(\frac{1}{2n+1})$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n})^2 < (\frac{1}{2n+1})$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n})< (\frac{1}{\sqrt{(2n+1)}})$$......... (2)$
From $(1)$&$(2)$,we get
$\prod_{n=1}^n(\frac{3n-2}{3n-1})^2< (\frac{1}{\sqrt{(2n+1)}})$
Now,
$lim_{n\to \infty} (\frac{1}{\sqrt{(2n+1)}}) = 0$
Hence , $lim_{n\to \infty} \prod_{n=1}^n(\frac{3n-2}{3n-1})^2  = 0$ 
Hence by leibniz criterion, the given series $\sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ is convergent. 
Answer credit,  phara narai
A: Let's prove that
$$
a_n=\frac{1\cdot 4\,\,\cdot\,\, ...\,\,\cdot\,\,(3n-2)}{2\cdot 5\,\,\cdot\,\, ...\,\,\cdot\,\,(3n-1)}
$$
converges to $0$. Clearly $a_n>0$ and since
$$
a_{n+1}=\frac{3n+1}{3n+2}a_n=\left(1-\frac{1}{3n+2}\right)a_n
$$
$a_n$ is decreasing. Hence, it is convergent. Suppose it does not converge to $0$. Then $a_n>C>0$ for all $n$. But: 
$$
a_{n+1}=a_1+\sum_{k=1}^n(a_{k+1}-a_k)=a_1-\sum_{k=1}^n\frac{1}{3n+2}a_k
$$
Since we have:
$$
\sum_{k=1}^n\frac{1}{3n+2}a_k>C\sum_{k=1}^n\frac{1}{3n+2}
$$
and the sum over $\frac{1}{3n+2}$ diverges, we get that for large enough $n$: $a_{n+1}<0$, a contradiction. Therefore $\lim_{n\rightarrow\infty}a_n=0$ as desired. Since $a_n$ is decreasing to $0$ and positive, the same holds for $a_n^2$ and you can apply Leibniz's test.
A: It is enough to prove that $\frac{\Gamma(n+1/3)}{\Gamma(n+2/3)}$ is decreasing to zero, then invoke Leibiz' criterion. On the other hand
$$\frac{\Gamma(n+1/3)}{\Gamma(n+2/3)}=\frac{1}{\Gamma(1/3)}B(n+1/3,1/3)=\frac{3}{\Gamma(1/3)}\int_{0}^{1}\color{red}{x^{3n}}(1-x^3)^{-2/3}\,dx$$
is obviously decreasing to zero: for any fixed $x\in(0,1)$, $x^n\searrow 0$ as $n\to +\infty$.
