Suppose $x, k > 1$. Is it true that,
$\displaystyle\sum_{j=0}^{\infty} \left( \cfrac{(k-1)(kx^2 - 2j(2j+1))}{(x+2j)(x+2j+1)(kx+2j)(kx+2j+1)} \right) > 0$?
This seems to be true for any values that I use, but I was wondering whether it could be shown formally? Thanks!
As you wrote it, I think that the expression is always negative. $$F(k,x)=\sum_{j=0}^{\infty} \left( \cfrac{\color{red}{-}(k-1)(kx^2 - 2j(2j+1))}{(x+2j)(x+2j+1)(kx+2j)(kx+2j+1)} \right) $$
reduces to $$2F(k,x)=k \left(\psi ^{(0)}\left(\frac{k x+1}{2}\right)- \psi ^{(0)}\left(\frac{k x}{2}\right)\right)-\left(\psi ^{(0)}\left(\frac{x+1}{2}\right)-\psi ^{(0)}\left(\frac{x}{2}\right)\right)$$ where appears the digamma function since, using partial fraction decomposition, the summand $$ \cfrac{(k-1)(kx^2 - 2j(2j+1))}{(x+2j)(x+2j+1)(kx+2j)(kx+2j+1)} $$ reduces to $$k\left(\frac{1}{2 j+1+k x}-\frac{1}{2 j+k x}\right)+\left(\frac{1}{2 j+x}-\frac{1}{2 j+1+x}\right)$$
For large values of $x$, the asymptotics is $$F(k,x)=-\frac{1-\frac{1}{k}}{4 x^2}+\frac{1-\frac{1}{ k^3}}{8x^4}-\frac{1-\frac{1}{k^5}}{4 x^6}+O\left(\frac{1}{x^8}\right)$$
May be, the minus sign in the numerator is a typo.
Edit
Since you changed the original minus sign to a plus sign now, change the sign in my answer and this makes you claim true.