Computing $\lim_{n\to \infty}{\frac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}}$ Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I.) In the first step I studied monotony:
$a_{n+1}-a_{n}=\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\cdot\cfrac{-2}{4n+7}<0$, $\{a_n\}$ is decreasing.
II.) In the second step I studied boundary.
$$1>a_{1}=\cfrac{5}{7}>a_{2}=\cfrac{45}{77}>\dots>a_{n}>0$$
III.) In the last step I know that $\{a_n\}$ converges to $a\in\mathbb R$.
$$a_{n+1}=a_{n}\cdot\cfrac{4n+1}{4n+3}$$
Taking the limit as $n\to\infty$:
$$a=a$$
No conclusion.
But if I apply Cesaro-Stolz?
IV.) Let $\{x_n\}_{n\ge1}^{\infty}=\{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)\}$ and $\{y_n\}_{n\ge1}^{\infty}=\{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)\}$. Then
$$\lim_{n\to \infty}{\cfrac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=\lim_{n\to \infty}{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)^2}{7\cdot11\cdot15\cdot\dots.\cdot(4n+5)}=?}}$$
If you have a simple solution, I would appreciate it. Thank you!
 A: Using the relation $x\Gamma(x)=\Gamma(x+1)$, we get
$$
\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac74\right)}\frac{\frac54\frac94\frac{13}4\cdots\frac{4n+1}4}{\frac74\frac{11}4\frac{15}4\cdots\frac{4n+3}4}
=\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}
$$
Therefore,
$$
\frac{5\cdot9\cdot13\cdots(4n+1)}{7\cdot11\cdot15\cdots(4n+3)}
=\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\frac{\Gamma\left(\frac74\right)}{\Gamma\left(\frac54\right)}
$$
Using Gautschi's Inequality, we get
$$
\left(\frac{4n+7}4\right)^{-1/2}\le\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\le\left(\frac{4n+3}4\right)^{-1/2}
$$
By the Squeeze Theorem, we get that the limit is $0$.
A: One can check that
$$
\bigg( \frac{4n+1}{4n+3} \bigg)^2 < \frac{n+1}{n+2}
$$
for all $n\ge0$. Therefore
$$
0 < a_n = \frac57 \frac9{11} \cdots \frac{4n+1}{4n+3} < \bigg( \frac23 \frac34 \cdots \frac{n+1}{n+2} \bigg)^{1/2} = \sqrt{\frac2{n+2}},
$$
and so $a_n\to0$ by the squeeze theorem.
A: We know that infinite products of the form $\displaystyle\prod_k(1+a_k)$ have the same nature as infinite series of the form $\displaystyle\sum_ka_k$ . Here, $a_k = \frac{4k+1}{4k+3}-1 = -\frac2{4k+3}$, which has order of magnitude $\frac1k$. Therefore the product diverges, having the same nature as the harmonic series. Also, an infinite product is said to diverge to $0$, which is clearly the case here, since each term is lesser than $1$.
