Is this inequality true? And can it be shown? If $x$ and $y$ are both positive, odd integers, greater than $1$, is it true that,
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > 0$?
It seems to be true, but I am having difficulty showing it for all relevant $x, y >1$.
 A: By multiplying and assembling the sums
$(y-1) \left( \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) - \ln(4) \right) + \displaystyle\sum_{i=\frac{x+1}{2}}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) =  $
$= \displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) - \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right)- (y-1)\ln(4) >0  \implies $
$\implies \displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right)+ (y-1)\ln(4).$
Since  $\sum_{i=1}^{\infty} \left( \frac{1}{i(2i-1)} \right) = \ln(4),$ we have
$\displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y}{i(2i-1)} \right) > \sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right) + \sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{y-1}{i(2i-1)} \right).$
Hence,
$\displaystyle\sum_{i=1}^{\frac{yx-1}{2}} \left( \frac{1}{i(2i-1)} \right) > \displaystyle\sum_{i=1}^{\frac{x-1}{2}} \left( \frac{1}{i(2i-1)} \right).$
