The Double Basel Problem

Question

I have been playing with the series which I had been calling the 'Double Basel problem' for the past couple of hours
$$ \sum_{n=1}^{\infty} \sum_{m=1}^\infty \frac{1}{{n^2 +m^2}}. $$ After wrestling with this for awhile, I managed to generalize a result demonstrated here. This identity is: $$ \sum_{m=1}^{\infty}\frac{1}{x^2+m^2} = \frac{1}{2x}\left[ \pi \coth{\pi x} - \frac{1}{x}\right]. $$ Hence the original series becomes: $$ \sum_{n=1}^{\infty} \frac{1}{2n}\left[\pi \coth{\pi n} - \frac{1}{n} \right]. $$

I have no idea where to go next with this problem. I seriously doubt that this series is convergent; however, I have been unable to prove it.

1. Can you prove that this series is divergent?
2. If it converges what is its value?

Thanks so much!

Answer 1

The series is monotonic in each argument, hence can be bounded below by an integral, of the form:

$$\int_1^\infty\int_1^\infty \frac{1}{x^2+y^2}dxdy=\int_0^{\pi/2}\int_1^\infty \frac{1}{r^2}rdrd\theta=\infty,$$

so the series diverges.

Answer 2

Less slick answer: Since $\displaystyle\lim_{x\to\infty}\coth(x)=1$ and since $\coth(x)\geq 1$ for $x\geq 0,$ then

$$\sum_{n=1}^{\infty} \frac{1}{2n}\left[\pi - \frac{1}{n} \right]=\sum_{n=1}^{\infty} \frac{\pi}{2n}-\frac{\pi}{2n^2} \leq \sum_{n=1}^{\infty} \frac{1}{2n}\left[\pi \coth{\pi n} - \frac{1}{n} \right],$$ and so the series diverges as $\sum_{n=1}^{\infty} \frac{\pi}{2n}$ is divergent and $\sum_{n=1}^\infty\frac{\pi}{2n^2}=\frac{\pi^3}{12}.$

Answer 3

Another proof that the series diverges: $$ \sum_{n=1}^{\infty} \sum_{m=1}^\infty \frac{1}{{n^2 +m^2}} \ge \sum_{n=1}^{\infty} \sum_{m=1}^n \frac{1}{{n^2 +m^2}} \ge \sum_{n=1}^{\infty} \sum_{m=1}^n \frac{1}{{n^2 +n^2}} = \sum_{n=1}^{\infty} \frac1{2n} = \infty. $$