# Interchanging the order of summation for a particular double series.

I suspect, based on numerical approximation, that

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^m}{2m+1}\frac{1}{n^{2m+1}} = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^m}{2m+1}\frac{1}{n^{2m+1}}= \sum_{m=1}^{\infty}\frac{(-1)^m}{2m+1}\zeta(2m+1).$$

Note that the double series fails to converge absolutely. We have absolute convergence if and only if each of the following integrals is convergent:

$$\int_1^{\infty} \int_1^{\infty}\frac{1}{2x+1} y^{-2x-1}\, dx \,dy, \\ \int_1^{\infty}\frac{1}{3} y^{-2x-1} \, dy, \\ \int_1^{\infty}\frac{1}{2x+1} \, dx ,$$

but the third is clearly divergent.

How can we justify switching the order of summation?

Separate the term for $n=1$:
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^m}{2m+1}\frac{1}{n^{2m+1}}=\frac\pi4-1+\sum_{n=2}^{\infty}\sum_{m=1}^{\infty}\frac{(-1)^m}{2m+1}\frac{1}{n^{2m+1}}\;.$$