Limit involving an arithmetic progression

Question

I have the following problem:

Let $(a_n)_{n\geq1}$ be an arithmetic progression such that $0<a_1<a_2$. Find the limit of the sequence $$y_n=\frac{a_1}{a_2}\cdot\frac{a_3}{a_4}\cdot...\cdot\frac{a_{2n-1}}{a_{2n}}$$

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 1

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.

Answer 2

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 $0 < a_1 \leq d$.

Answer 3

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 $z_n=y_n \sqrt n$

Answer 4

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 $-\infty$.