Question: I have the following problem:

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> Let $(a\_n)\_{n\geq1}$ be an arithmetic progression such that $0<a\_1<a\_2$. Find the limit of the sequence $$y\_n=\frac{a\_1}{a\_2}\cdot\frac{a\_3}{a\_4}\cdot...\cdot\frac{a\_{2n-1}}{a\_{2n}}$$ 
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It is not complicated to see that $y\_n$ is decreasing and bounded, so it is convergent, but I can't find it's limit. It is clearly less than $\frac{a\_1}{a\_2}$. I tried using the AM-GM inequality but I obtained something trivial.

Answer: 
We have $\frac{a\_{k}}{a\_{k+1}} < \frac{a\_{k+1}}{a\_{k+2}}$ for all $k$, so

$$y\_n^2 < \biggl(\frac{a\_1}{a\_2}\cdot\frac{a\_2}{a\_3}\biggr)\cdot\biggl(\frac{a\_3}{a\_4}\cdot\frac{a\_4}{a\_5}\biggr) \cdot\dotsc \cdot\biggl(\frac{a\_{2n-1}}{a\_{2n}}\cdot\frac{a\_{2n}}{a\_{2n+1}}\biggr) = \frac{a\_1}{a\_{2n+1}}\,.$$

Thus

$$0 < y\_n < \sqrt{\frac{a\_1}{a\_{2n+1}}} \to 0\,.$$

Squeezing finishes it.
*Too long for a comment.*

Using $a\_n=A+(n-1)B$ and Pochhammer symbols the numerator write
$$2^{n-1}A\, B^{n-1} \left(\frac{A+2B}{2 B}\right)\_{n-1}$$ and the denominator
$$2^{n-1} (A+B) B^{n-1} \left(\frac{A+3 B}{2 B}\right)\_{n-1}$$ making
$$y\_n=\frac{A}{A+B} \frac{ \left(\frac{A+2B}{2 B}\right)\_{n-1}}{ \left(\frac{A+3 B}{2
 B}\right)\_{n-1}}$$ Using asymptotics
$$y\_n=\frac{A}{A+B} \frac{ \Gamma \left(\frac{A+3 B}{2 B}\right)}{ \Gamma
 \left(\frac{A+2B}{2 B}\right)}\frac 1{\sqrt n}+O\left(\frac{1}{n^{3/2}}\right)$$ What would have very interesting is to compute the limit of $z\_n=y\_n \sqrt n$