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: 
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 $-\infty$.
The limit is probably $0$. 

In the case where $a\_1=1$ and the arithmetic progression is $1$, we have 
\begin{align\*}
y\_n & =\frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4 \cdots 2n} \\
 & = \frac{(2n)!}{2^{2n}(n!)^2}
\end{align\*}
With the Stirling formula $n!\sim \sqrt{2\pi n}\,\left({n \over {\rm e}}\right)^n$, it follows
$$ y\_n \sim \frac{1}{\sqrt{\pi n}}$$
So, in this particular case $y\_n$ tends to zero...

In the general case 
$$ y\_n = \prod\_{k=1}^n\frac{a\_1+(2k-2)d}{a\_1+(2k-1)d} := \prod\_{k=1}^n\frac{1+(2k-2)u}{1+(2k-1)u} $$
(with $u:=\frac{d}{a\_1}$)

But (I assume $u, u'>0$, if not, reverse the final inequality)
\begin{align\*}
& \frac{1+(2k-2)u}{1+(2k-1)u} \leq \frac{1+(2k-2)u'}{1+(2k-1)u'} \\
\Leftrightarrow \quad & (1+(2k-2)u)(1+(2k-1)u') \leq (1+(2k-2)u')(1+(2k-1)u) \\
\Leftrightarrow \quad & (2k-2)u+(2k-1)u'\leq (2k-2)u'+(2k-1)u \\
\Leftrightarrow \quad & u'\leq u
\end{align\*}

So, the first particular case gives already that $y\_n$ tends to $0$ if $1\leq \frac{d}{a\_1}$, or $0 < a\_1 \leq d$.