\(\displaystyle \dfrac{1}{4 * 1 + 1} = \dfrac{1}{5}. \text { And } \left ( \dfrac{1}{2} \right )^2 = \dfrac{1}{4}. \text { And } \dfrac{1}{3 * 1 + 1} = \dfrac{1}{4}.\)
So the given inequality is not true for n = 1. Possibly the problem said n > 1.
\(\displaystyle \dfrac{1}{4 * 2 + 1} = \dfrac{1}{9} = \dfrac{448}{4032}. \text { And } \left ( \dfrac{1 * 3}{2 * 4} \right )^2 = \dfrac{9}{64} = \dfrac{567}{4032}.\)
\(\displaystyle \text { And } \dfrac{1}{3 * 2 + 1} = \dfrac{1}{7} = \dfrac{576}{4032}.\)
So it is true for n = 2.
So the first thing I would do if I wanted to try a proof by induction is to restate the problem.
\(\displaystyle n \in \mathbb Z^+ \implies \dfrac{1}{4n + 1} < \displaystyle \left \{ \left ( \prod_{j=1}^n 2j - 1 \right ) \div \left ( \prod_{j=1}^n 2j \right ) \right \}^2 \le \dfrac{1}{3n + 1}.\)
We know that is true if n = 1. Therefore
\(\displaystyle \exists \text { non-empty set } \mathbb S \text { such that } k \in \mathbb S \implies k \in \mathbb Z^+ \text { and}\)
\(\displaystyle \dfrac{1}{4k + 1} < \displaystyle \left \{ \left ( \prod_{j=1}^k 2j - 1 \right ) \div \left ( \prod_{j=1}^k 2j \right ) \right \}^2 \le \dfrac{1}{3k + 1}.\)
Now what I would try to show is
\(\displaystyle \dfrac{1}{4(k + 1) + 1} < \displaystyle \left \{ \left ( \prod_{j=1}^{k+1} 2j - 1 \right ) \div \left ( \prod_{j=1}^{k+1} 2j \right ) \right \}^2 < \dfrac{1}{3(k + 1) +1}.\)
If I could do that, I would prove the given theorem because \(\displaystyle < \implies \le\) and have the basis for the more restrictive theorem for n > 1.
EDIT: An alternate approach, still by induction, is to prove:
\(\displaystyle n \in \mathbb Z^+ \implies \dfrac{1}{4n + 5} < \displaystyle \left \{ \left ( \prod_{j=1}^{n+1} 2(j - 1) \right ) \div \left ( \prod_{j=1}^{n+1} 2j \right ) \right \}^2 < \dfrac{1}{3n + 4}.\)
