Magical Summation Interchange In this answer I use the following result:
$$\lim_{p\to\infty}\sum_{n=1}^p\sum_{k=0}^\infty\frac{(-1)^{n+k}}{n(2k+1)^n} = \lim_{p\to\infty}\sum_{k=0}^p\sum_{n=1}^\infty\frac{(-1)^{n+k}}{n(2k+1)^n}$$
This came purely by intuitition though - I have no idea how to prove this result!
I have numerically checked this and it seems to hold; moreover, using this interchange I derived the correct result in the linked post. If I were simply interchanging the summations I could produce a trivial proof, though I am also interchanging summation limits.   
Is there an easy proof of this?
 A: You may notice that
$$ \int_{0}^{1} x^{2k}\frac{(-\log x)^{n-1}}{(n-1)!}\,dx = \frac{1}{(2k+1)^n} $$
hence both the RHS and LHS equal
$$ \int_{0}^{1} \sum_{k\geq 0}\sum_{n\geq 1}\frac{(-1)^n(-\log x)^{n-1}}{n!}(-1)^k x^{2k}\,dx =\int_{0}^{1}\frac{1-x}{\log(x)}\cdot\frac{1}{1+x^2}\,dx$$
by the dominated convergence theorem, since the exponential function is an entire function and $\frac{1}{1+z^2}$ is an analytic function whose radius of convergence at the origin equals one. Through the substitution $x=e^{-t}$ and Frullani's theorem the last integral equals $$\log\,\left(\frac{\pi\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}\right)=\log\,\left(\frac{\Gamma\left(\frac{3}{4}\right)^2}{\sqrt{2\pi}}\right)\approx -0.512376630342 $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\lim_{p \to \infty}\sum_{n = 1}^{p}
\sum_{k = 0}^{\infty}{\pars{-1}^{n + k} \over n\pars{2k + 1}^{n}} & =
\lim_{p \to \infty}\sum_{n = 1}^{p}{\pars{-1}^{n} \over n}
{1 \over \ic}\sum_{k = 0}^{\infty}{\ic^{2k + 1} \over \pars{2k + 1}^{n}}
\\[5mm] &=
\lim_{p \to \infty}\sum_{n = 1}^{p}{\pars{-1}^{n} \over n}\,
{1 \over \ic}\sum_{k = 1}^{\infty}{\ic^{k} \over k^{n}}\,
{1 - \pars{-1}^{k} \over 2} =
\lim_{p \to \infty}\Im\sum_{n = 1}^{p}{\pars{-1}^{n} \over n}\,
\sum_{k = 1}^{\infty}{\ic^{k} \over k^{n}}
\\[5mm] &=
\lim_{p \to \infty}\Im\sum_{k = 1}^{\infty}\ic^{k}
\sum_{n = 1}^{p}{\pars{-1/k}^{n} \over n} =
-\Im\sum_{k = 1}^{\infty}\ic^{k}\ln\pars{1 + {1 \over k}}
\\[5mm] & =
-\,\Im\sum_{k = 1}^{\infty}\ic^{k}\int_{0}^{1}{\dd t \over t + k} =
-\int_{0}^{1}\Im\sum_{k = 1}^{\infty}{\ic^{k} \over k + t}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 1/2 + t/2}\,\dd t
\\[5mm] & =
{1 \over 4}\int_{0}^{1}
\sum_{k = 0}^{\infty}\pars{{1 \over k + 3/4 + t/4} -
{1 \over k + 1/4 + t/4}}\,\dd t
\\[5mm] & =
{1 \over 4}\int_{0}^{1}\bracks{\Psi\pars{{t \over 4} + {1 \over 4}} -
\Psi\pars{{t \over 4} + {3 \over 4}}}\,\dd t\quad
\pars{~\substack{\Psi:\ Digamma\ Function}~}
\\[5mm] & =
\left.\ln\pars{\Gamma\pars{t/4 + 1/4} \over \Gamma\pars{t/4 + 3/4}}
\right\vert_{\ 0}^{\ 1} =
\ln\pars{{\Gamma\pars{1/2} \over \Gamma\pars{1}}
\,{\Gamma\pars{3/4} \over \Gamma\pars{1/4}}}
\\[5mm] & =
\ln\pars{\root{\pi}
\,{\Gamma^{\,2}\pars{3/4} \over \Gamma\pars{3/4}\Gamma\pars{1/4}}} =
\ln\pars{\root{\pi}\Gamma^{\,2}\pars{3/4} \over \pi/\sin\pars{\pi/4}}
\\[5mm] & =
\bbx{\ln\pars{\Gamma^{\,2}\pars{3/4} \over \root{2\pi}}} \approx -0.5123
\end{align}
