In my integration adventures, I came across this sum which I could not simplify:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(2n+1)}{2n+1}$$
Wolfram seems to believe the sum diverges and is not of much help here.
Does a closed form for this sum exist? If not, can this sum be transformed nicely that has faster convergence?
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: $$ \sum_{n=0}^\infty (-1)^n \frac{\log(2n+1)}{2n+1} = - \frac{\pi}{4} \left( \gamma + \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \right) \approx -0.1929013\color\gray{167969124} $$
s approaches in the limit to solve your problem?
– Sasha
Jun 28 '22 at 17:35
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.
I get (modulo algebra errors) that it does converge.
Pair the odd and even terms:
If $t(n) =\frac{(-1)^{n}\log(2n+1)}{2n+1} $, then
$\begin{array}\\ t(2n-1)+t(2n) &=\frac{(-1)^{2n-1}\log(2(2n-1)+1)}{2(2n-1)+1}+\frac{(-1)^{2n}\log(2(2n)+1)}{2(2n)+1}\\ &=\frac{-\log(4n-1)}{4n-1}+\frac{\log(4n+1)}{4n+1}\\ &=\frac{(4n-1)\log(4n+1)-(4n+1)\log(4n-1)}{(4n-1)(4n+1)}\\ &=\frac{(4n-1)(\log(4n)+\log(1+1/(4n)))-(4n+1)(\log(4n)+\log(1-1/(4n)))}{(4n-1)(4n+1)}\\ &=\frac{-2\log(4n)+4n(\log(1+1/(4n))-\log(1-1/(4n))-\log(1+1/(4n))-\log(1-1/(4n))}{(4n-1)(4n+1)}\\ &=\frac{-2\log(4n)+4n(1/(4n)-1/(32n^2)+O(1/n^3)+(1/(4n)+1/(32n^2)+O(1/n^3))-\log(1-1/(16n^2))}{(4n-1)(4n+1)}\\ &=\frac{-2\log(4n)+4n(1/(2n)+O(1/n^3))+1/(16n^2)+O(1/n^4)}{(4n-1)(4n+1)}\\ &=\frac{-2\log(4n)+2+O(1/n^2)}{(4n-1)(4n+1)}\\ \end{array} $
and the sum of these converge.
