How do I calculate $\frac{\ln(2)}{\ln(3)}\frac{\ln(4)}{\ln(5)}...\frac{\ln(2n)}{\ln(2n+1)}$? The product has only positive factors so it has zero as lower bound. Also the product is decreasing as all its factors are less than one. In conclusion the series must have a limit. I also compute the first 150 values of the product and I got around 0.297. I believe that the product converges very, very, slowly to zero, but I can't prove it.
 A: The given product equals
$$\exp\sum_{n\geq 1}\log\frac{\log(2n)}{\log(2n+1)}=\exp\sum_{n\geq 1}-\int_{2n}^{2n+1}\frac{dx}{x\log x}\\=\exp\left[C-\sum_{n\geq 2}\frac{1}{(2n+\eta_n)\log(2n+\eta_n)}\right]$$
with $\eta_n\in(0,1)$ by the mean value theorem. Since both the series $\sum_{n\geq 2}\frac{1}{(2n+1)\log(2n+1)}$ and $\sum_{n\geq 2}\frac{1}{2n\log(2n)}$ are divergent by Cauchy's condensation test, the given product is convergent to zero, albeit very slowly.
A: Hint:
Since
$$\ln(1+x)\sim_0x$$
It follows from logarithmic rules that
$$\frac{\ln(2n+1)}{\ln(2n)}=1+\frac{\ln(1+1/2n)}{\ln(2n)}\sim_\infty1+\frac1{2n\ln(2n)}$$
Reciprocate the product, then take the log of it, and with this, we find
$$-\ln\prod_{k=1}^\infty\frac{\ln(2k)}{\ln(2k+1)}\sim\sum_{k=1}^\infty\frac1{2n\ln(2n)}=+\infty$$
Thus, the given product should go to
$$\prod_{k=1}^\infty\frac{\ln(2k)}{\ln(2k+1)}=e^{-\infty}=0$$
This approach may be rigorously done using squeezing and the bounds
$$x\ln(2)<\ln(1+x)<x\quad\forall x\in(0,1)$$
A: $$\ln \left(\dfrac{\ln(2n)}{\ln(2n+1)}\right) \sim \ln\left(1 - \frac{1}{2 n \ln(n)}\right) \sim - \frac{1}{2 n \ln(n)}$$
and $\sum_n \frac{1}{n \ln(n)}$ diverges, so the limit is $0$.
A: I just read this question and basically we can use the same approach.
If we denote
$$P_n=\prod_{i=1}^n \frac{ln(2i)}{ln(2i+1)} \ and \ Q_n=\prod_{i=1}^n \frac{ln(2i+1)}{ln(2i+2)} \ then \  P_n*Q_n = \frac{ln(2)}{ln(2n+2)} \, now \ Q_n \ge P_n \ge 0 \\
thus
\ 0 \le P_n \le \sqrt{\frac{ln(2)}{ln(2n+2)}}$$ 
