Showing $\lim_{x\rightarrow 1^-} \sum_{k=0}^\infty (-x)^k (2k+1) \ln(2k+1) = \frac{-2C}{\pi}$

Question

While I was playing around with divergent summation, I noticed that the following appears to be true:

$$\lim_{x\rightarrow 1^-} \sum_{k=0}^\infty (-x)^k (2k+1) \ln(2k+1) = \frac{-2C}{\pi}$$

where $C = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}$ is Catalan's constant. How can I show this?

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I'll note that the above is Abel summation, and so different choices of divergent summation should work; for example, I suspect:

$$\lim_{n \to \infty} \sum_{k=0}^{n}\frac{\binom{2n}{n+k}}{\binom{2n}{n}} (-1)^k (2k+1)\ln(2k+1) = \frac{-2C}{\pi} $$

The reason I mention this latter limit is because I've seen some similar limits performed with the factor of $\binom{2n}{n+k}$, and so I hope that this might make the problem tractable (see also this question and the comments and answer). I'll accept an answer that proves either of the above limits, since I suspect them to be equivalent.

Answer 1

For $0<x<1$ we have $f(x):=\sum_{k=0}^\infty(2k+1)(-x)^k=(1-x)/(1+x)^2$.

Using Frullani integral $\log z=\int_0^\infty(e^{-t}-e^{-zt})\,dt/t$, we obtain $$F(x):=\sum_{k=0}^\infty(-x)^k(2k+1)\log(2k+1)=\int_0^\infty\big(f(x)-f(xe^{-2t})\big)\frac{e^{-t}}{t}\,dt$$ (still for $0<x<1$); the change $\sum\int\mapsto\int\sum$ is justified by absolute convergence. Moreover, it is easy to check that $(a,x)\mapsto\big(f(x)-f(ax)\big)/(1-a)$ is bounded on $0<a,x<1$, so that we may take $x\to1^-$ under the integral sign (by dominated convergence): $$F(1^-)=-\int_0^\infty f(e^{-2t})\frac{e^{-t}}{t}\,dt=-\frac12\int_0^\infty\frac{\sinh t}{\cosh^2 t}\frac{dt}{t}.$$ Now, using the following integral (say, a special case of the beta function) $$g(x):=\int_0^\infty\frac{\cosh 2xt}{\cosh^2 t}\,dt=\frac{x\pi}{\sin x\pi}\qquad(|x|<1)$$ we get $F(1^-)=-\int_0^{1/2}g(x)\,dx$, giving the result by $2C=\int_0^{\pi/2}\frac{x\,dx}{\sin x}$.