# 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?

• Since the family of terms where $n \geqslant 2$ and $k \geqslant 1$ is absolutely summable, one needs only care about $n = 1$ or $k = 0$. Splitting these off, you get the result straightforwardly. Jun 27, 2017 at 18:15

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 $$\log\,\left(\frac{\pi\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}\right)=\log\,\left(\frac{\Gamma\left(\frac{3}{4}\right)^2}{\sqrt{2\pi}}\right)\approx -0.512376630342$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}