Is the series convergent Is series $\sum_1^\infty \frac{\ln(1+1/2) \ln(1+1/4) \cdots \ln(1+1/(2n))}{\ln(1+1/1) \ln(1+1/3) \cdots \ln(1+1/(2n-1))} = \sum_{n=1}^\infty \prod_{m=1}^n \ln(1+1/(2m))/(\ln(1+1/(2m-1))$ convergent ?
 A: Hint. One may use, for $x \in [0,1]$,
$$
x-\frac{x^2}2 \leq\ln(1+x)\leq x \tag1
$$ giving, as $n \to \infty$,
$$
\begin{align}
\prod_{k=1}^n\frac1{2k}\prod_{k=1}^n\left(1-\frac1{4k}\right)\leq \ln(1+1/2) \ln(1+1/4) \cdots \ln(1+1/(2n)) \tag2
\end{align}
$$ and 
$$
\begin{align}
\prod_{k=1}^n(2k-1)\leq \frac1{\ln(1+1/3) \ln(1+1/5) \cdots \ln(1+1/(2n-1))} \tag3
\end{align}
$$ and
$$
\prod_{k=1}^n\frac{2k-1}{2k}\prod_{k=1}^n\left(1-\frac1{4k}\right)\leq\frac{\ln(1+1/2) \ln(1+1/4) \cdots \ln(1+1/(2n))}{\ln(1+1/3) \ln(1+1/5) \cdots \ln(1+1/(2n-1))}. \tag4
$$
One may use, the Euler gamma function to reduce the left hand side and, as $n \to \infty$, one gets
$$
\prod_{k=1}^n\frac{2k-1}{2k}\prod_{k=1}^n\left(1-\frac1{4k}\right)=\frac {(2n)!} {4^n (n!)^2}\times \frac{\Gamma\left(\frac34+n\right)}{\Gamma\left(\frac34\right) n!} \sim \frac1{\sqrt{\pi }\: \Gamma\left(\frac34\right) }\times \frac{1}{n^{3/4}} \tag5
$$ giving, by comparison, the divergence of the initial series.
A: We have $u - u^2/2 \le \ln (1+u)\le u$ for $u\ge 0.$ Thus for $x\ge 1,$
$$ \frac{\ln (1+1/(2x))}{\ln (1+1/(2x-1))} \ge \frac{1/(2x) - 1/(8x^2)}{1/(2x-1))} = 1-3/(4x) + 1/(8x^2) \ge 1-3/(4x).$$
The $n$th  term of our series is thus at least
$$ P_n =  \prod_{k=1}^n (1-3/(4k)).$$
To estimate $P_n, $ apply $\ln $ and use the inequality $\ln (1-u) \ge -u - u^2,$ which holds for $0< u < 1/2.$ Thus
$$\ln P_n \ge \ln (1/4) + \sum_{k=2}^{n}\ln (1-3/(4k)) \ge \ln (1/4) + \sum_{k=2}^{n} [-3/(4k) - 9/(16k^2)] $$ $$ \ge \ln (1/4) -(3/4)\ln n -\sum_{k=2}^{\infty}9/(16k^2).$$
Exponentiating back shows there is a constant $c> 0,$ independent of $n,$ such that $P_n \ge c/n^{3/4}.$ Since $\sum 1/n^{3/4} = \infty,$ our series diverges.
