What would be an equivalent to the ratio of all odd numbers by all even numbers? Consider the sequence U defined by:

*

*$U_0 = 1$

*$U_{n+1} = U_{n}*(2n+1)/(2n+2)$
I am interested in how this sequence behaves near infinity.
As the terms are multiplied, each step, by $(2n+1)/(2n+2)$, which is in the $]0,1[$ interval, it should converge, probably toward 0 although I don't know how to prove that. Knowing the limit is not enough though: I would ideally be able to have a polynomial equivalent for how fast it converges (or an other simple equivalent if polynomial is not possible).
 A: You want $\prod (1-\frac{1}{2n+2})$.  This converges to a non-zero number if and only if $\sum \ln (1-\frac{1}{2n+2})$ converges.  But the Taylor series for $\ln (1-x)$ shows us that this sum is greater (in absolute value) than $\frac 12$ of the harmonic series, so the product diverges to $0$.
A: Robert Shore's answer is probably the best way to think about this problem. But just to give a very hands-on approach: note that
$$
\bigg( \frac{2n+1}{2n+2} \bigg/ \sqrt{\frac{n+1}{n+2}} \biggr)^2 = \frac{(2n+1)^2(n+2)}{(2n+2)^2(n+1)} = 1 - \frac{3n+2}{4(n+1)^3} < 1
$$
for all $n\ge0$. In particular, $\displaystyle\frac{2n+1}{2n+2} < \sqrt{\frac{n+1}{n+2}}$ always, and so
$$
0 < \prod_{n=0}^N \frac{2n+1}{2n+2} < \sqrt{\prod_{n=0}^N \frac{n+1}{n+2}} = \sqrt{\frac1{N+2}},
$$
which is enough to imply that $\displaystyle\prod_{n=0}^\infty \frac{2n+1}{2n+2}$ diverges to $0$.
(Indeed, changing the $\displaystyle\frac{n+1}{n+2}$ to $\displaystyle\frac{n}{n+1}$ produces the lower bound $\displaystyle\sqrt{\frac1{4(N+1)}}$ for the partial product, so $N^{-1/2}$ is the right order of decay.)
