Prove that : A < 1/5 < B for a Product of Fractions Sequence

Question

Consider the numbers A and B such that: $$A = \frac{1}{2}\cdot\frac{3}{4} \cdots \frac{23}{24}$$ $$B = \frac{2}{3}\cdot\frac{4}{5} \cdots \frac{24}{25}$$ Prove that $A < 1/5 < B$. My approach: honestly, I don't have any idea how I can compare $A$ and $B$, but I could show that one of them is less than 1/5 and the other is greater than 1/5. And that's by simply realizing that if $x < y$, then $x < \sqrt{xy} < y$, which means that if $A < B$, then $A < \sqrt{AB} < B$. From here, I tried to compute $AB$ as follows: $$AB = \frac{1 \cdot 3 \cdot 5 \cdots 23}{2 \cdot 4 \cdot 6 \cdots 24} \cdot \frac{2 \cdot 4 \cdot 6 \cdots 24}{3 \cdot 5 \cdot 7 \cdots 25}$$ All terms cancel out, and I conclude that $AB = 1/25$, hence $\sqrt{AB} = 1/5$, hence $A < 1/5 < B$ if and only if $A < B$, which I couldn't show, unfortunately.

Answer 1

The piece of information you need can be found by realizing that both $A$ and $B$ are a product of 12 positive fractions, such that the ith fraction in $A$ is always smaller than the ith fraction in $B$. Hence, $A<B$.

Answer 2

Hint: We have $0<\frac12<\frac23$, $0<\frac34<\frac45$, ..., $0<\frac{23}{24}<\frac{24}{25}$ so ...