# 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 $$\log\,\left(\frac{\pi\sqrt{2\pi}}{\Gamma\left(\frac{1}{4}\right)^2}\right)=\log\,\left(\frac{\Gamma\left(\frac{3}{4}\right)^2}{\sqrt{2\pi}}\right)\approx -0.512376630342$$