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.

$(\frac{1}{2})^2 < \frac{1}{2}$,
$(\frac{4}{5})^2 < \frac{3}{4}$,
$(\frac{7}{8})^2 < \frac{5}{6}$, $\dots \dots$
$(\frac{3n-2}{3n-1})^2 < \frac{2n-1}{2n}$
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
