<s>[INST] Study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^25^2\cdots (4n-3)^2}{3^27^2\cdots(4n-1)^2}$ I need to study the convergence of the series $\sum_{n=1}^{\infty}\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}$.
Now, I think we can do it by using the fact that if we have a series $\sum_{n=1}^{\infty}a_n$ and we can find $b_n$ so that $a_n<b_n$ then: 
if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ i convergent 
or 
if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent
What I did so far is:
I said that $\frac{1}{3}<\frac{3}{5},\frac{5}{7}<\frac{7}{9},...,\frac{4n-3}{4n-1}<\frac{4n-1}{4n+1}$ 
and we get that 
$\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}<\frac{3*7*...*(4n-1)}{5*9*...*(4n+1)}$ 
and if we multiply both sides with $\frac{1*3*...*(4n-3)}{3*7*...*(4n-1)}$ 
we get that
$\frac{1^2*5^2*...*(4n-3)^2}{3^2*7^2*...*(4n-1)^2}<\frac{1}{4n-1}$ but that does not really help me.
 [/INST] We are interested in
$$ \sum_{n\geq 1}\left[\frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(n+\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(n+\frac{3}{4}\right)}\right]^2 $$
and by Gautschi's inequality the main term of this series behaves like $\frac{K}{n}$ as $n\to +\infty$, hence the given series is divergent by asymptotic comparison with the harmonic series. Namely $K=\frac{1}{2\pi^2}\Gamma\left(\frac{3}{4}\right)^4$.

Elementary alternative: you may prove by induction that for any $n\in\mathbb{N}^+$
$$\left(\prod_{k=1}^{n}\frac{4k-3}{4k-1}\right)^2 \geq \frac{1}{9n} $$
holds (since $\left(\frac{4n+1}{4n+3}\right)^2-\frac{n}{n+1}=\frac{1}{(n+1)(4n+3)^2}>0$) and the conclusion is just the same.
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