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.
The logarithm of the product is

$$\sum\_k\log\frac{a+2kr-r}{a+2kr}=\sum\_k\log\left(1-\frac r{a+2kr}\right)<-\sum\_k\frac r{a+2kr}$$

and the series is bounded by the harmonic one, the sum diverges to $-\infty$.