Help evaluate the following sum:
$$\sum_{n=1}^{\infty}\left(\frac{1}{4n-1}-\frac{1}{4n-1}\ln\left(2n-1\right)+\frac{4}{\left(4n-1\right)\left(4n-3\right)}\left(\psi^{(-2)}\left(2n-\frac{1}{2}\right)-\psi^{(-2)}\left(2n-1\right)\right)\right)$$
This arose from a relation to an integral.
We could try to find a form for the partial sums and take the limit. How would we go about that?
Update
I found an equivalent form as follows $$\frac{\pi}{8}\left(6\ln A+\ln\pi-\frac{1}{6}\ln2-1\right)+\sum_{n=0}^{\infty}\left(\frac{\ln\left(4n+2\right)}{\left(4n+1\right)}+\frac{2}{\left(4n+1\right)\left(4n+3\right)}\sum_{k=1}^{4n+2}\left(-1\right)^{\left(k+1\right)}k\ln\left(k\right)\right)$$