A new sum that equals $-\frac12 \ln (\frac{\pi}{2})$

(See EDIT)

I found this result while working another problem in two slightly different directions and it kind of took me by surprise.

$$\begin{align}&\sum_{n=1}^{\infty}\left(\frac{\ln4n}{4n-1}-\frac{2}{\left(4n-1\right)\left(4n+1\right)}\sum_{k=1}^{4n}\left(-1\right)^{k}k\ln k\right)\\=-\frac{1}{2}\ln\left(\frac{\pi}{2}\right)-&\sum_{n=1}^{\infty}\left(\frac{\ln\left(4n-2\right)}{4n-3}-\frac{2}{\left(4n-3\right)\left(4n-1\right)}\sum_{k=1}^{4n-2}\left(-1\right)^{k}k\ln k\right)\end{align}$$

The only difference between the summation expressions is a shift in the index $n$ by $1/2$.

1. Is there a neat way to prove it? (I feel that explaining the long process that brought me here would dilute the question. I wonder if there is a straight-forward solution starting with one of the sums.)
2. Are there other sums that have general ways to be expressed in terms of themselves with a non-integer shift of the index?
3. Extra: Can we find closed forms for the sums above?

EDIT:

After seeing the comments, I realized the obvious fact that my main question is equivalent to proving

$$\sum_{n=1}^{\infty}\left(\frac{\ln2n}{2n-1}-\frac{2}{\left(2n-1\right)\left(2n+1\right)}\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k\right)=-\frac{1}{2}\ln\left(\frac{\pi}{2}\right)$$

However I am also interested in finding a closed form for

$$\sum_{n=1}^{\infty}(-1)^n\left(\frac{\ln2n}{2n-1}-\frac{2}{\left(2n-1\right)\left(2n+1\right)}\sum_{k=1}^{2n}\left(-1\right)^{k}k\ln k\right)$$