Evaluating a slow sum

Solution 1:

Using $$\frac{\log(2n+1)}{2n+1} = -\lim_{s \to 1^+} \frac{\mathrm{d}}{\mathrm{d}s} \frac{1}{(2n+1)^s}$$ as well as absolute convergence of $\sum_{n=0}^\infty (-1)^n (2n+1)^{-s}$ for $s>1$ we get $$\begin{eqnarray} \sum_{n=0}^\infty (-1)^n \frac{\log(2n+1)}{2n+1} &=& -\lim_{s \to 1^+} \frac{\mathrm{d}}{\mathrm{d}s} \sum_{n=0}^\infty (-1)^n (2n+1)^{-s} \\ &=& -\lim_{s \to 1^+} \frac{\mathrm{d}}{\mathrm{d}s} \left(2^{-2 s} \left( \zeta\left(s,\frac{1}{4}\right) - \zeta\left(s,\frac{3}{4}\right) \right) \right) \end{eqnarray} $$ Using $\zeta(s,a) = \frac{1}{s-1} - \psi(a) + \gamma_1(a)(s-1) + \mathcal{o}(s-1)$, where $\psi(a)$ is the digamma function, and $\gamma_1(a)$ is the first generalized Stieltjes constant, we get: $$ \sum_{n=0}^\infty (-1)^n \frac{\log(2n+1)}{2n+1} = \frac{\pi}{2} \log(2) + \frac{1}{4} \left( \gamma_1\left(\frac{1}{4}\right) - \gamma_1\left(\frac{3}{4}\right) \right) $$ the same combination of generalized Stieltjes constants appeared in another answer of mine, leading to the following closed form for the sum: $$ \sum_{n=0}^\infty (-1)^n \frac{\log(2n+1)}{2n+1} = - \frac{\pi}{4} \left( \gamma + \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \right) \approx -0.1929013\color\gray{167969124} $$

Solution 2:

I would like to propose a slightly different approach.
For $\text{Re}(s)>0$, we may introduce the Dirichlet L-function $$\begin{eqnarray*} L(\chi_4,s)=\sum_{n\geq 1}\frac{\chi_4(n)}{n^s}&=&\sum_{k\geq 0}\left(\frac{1}{(4k+1)^s}-\frac{1}{(4k+3)^s}\right)\\&=&\prod_{p}\left(1-\frac{\chi_4(p)}{p^s}\right)^{-1} \tag{A}\end{eqnarray*}$$ and notice that the value of the given series just depends on $\frac{d}{ds}L(\chi_4,s)$ at $s=1$, i.e. $$ L'(\chi_4,1) = L(\chi_4,1)\cdot\frac{L'(\chi_4,1)}{L(\chi_4,1)}=\frac{\pi}{4}\sum_{p}\frac{\log p}{p\cdot\chi_4(p)-1}.\tag{B}$$ On the other hand, by the (inverse) Laplace transform we have $$ L(\chi_4,s) = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}\,dx}{2\cosh x}\tag{C}$$ hence: $$ L'(\chi_4,s) = \frac{\pi\gamma}{4}+\color{blue}{\int_{0}^{+\infty}\frac{\log x}{2\cosh x}\,dx} \tag{D}$$ and the whole problem boils down to the evaluation of the blue integral, which is clearly related to the Gudermannian function by integration by parts. According to Gradshteyn-Rizhyk 4.371 we actually have $$ \int_{0}^{+\infty}\frac{\log x}{2\cosh x}\,dx = \frac{\pi}{4}\,\log\,\left(\frac{4\pi^3}{\Gamma\left(\tfrac{1}{4}\right)^4}\right)\tag{E}$$ leading to an unexpected closed form for the RHS of $(B)$, too. I guess that a proof of $(E)$ can be derived from differentiating the reflection formula for the involved Dirichlet L-function.