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 $n=1$:

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

Now the third integral converges and the double series is absolutely convergent.

More generally, you may also be interested in this useful general theorem on the interchangeability of limits if the convergence to the inner limit is uniform with respect to the outer variable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.