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.

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

Thanks so much!

For large $n$, $\coth(n \pi) \sim 1$ so the first term behaves like $1/n$ which diverges. More precisely, $\coth(n \pi)/n >1/n$ for $n>0$ — Sal, Jun 22 '21 at 23:26
Inasmuch as $\lim_{n\to\infty}\coth(\pi n)=1$, then the series of interest diverges. — Mark Viola, Jun 22 '21 at 23:31
The series can be made more interesting by adding a factor $(-1)^n$ to get $\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^n}{m^2+ n^2}$ which is convergent and has a nice closed form. — Paramanand Singh, Jun 22 '21 at 23:36
Could you link to this series? I would love to see the closed form! Thanks. — guavas222, Jun 22 '21 at 23:44
The sum in my previous comment can be evaluated using the approach in this answer — Paramanand Singh, Jun 23 '21 at 03:53
Also see this answer : https://math.stackexchange.com/a/4113092/72031 — Paramanand Singh, Jun 23 '21 at 03:58

Alex R. · Accepted Answer · 2021-06-23T06:28:04.690

11

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.

edited Jun 23 '21 at 06:28

answered Jun 22 '21 at 23:33

Alex R.

32,771

Slick answer – Chickenmancer Jun 22 '21 at 23:35
Thanks for this, but the area you are integrating over isn't well preserved in the polar coordinates. Is the divergence still true? – guavas222 Jun 22 '21 at 23:41
@guavas222 Changing variable can't turn a divergent integral into a convergent one if the integrand is nonnegative. – Rivers McForge Jun 23 '21 at 00:05
2

@guavas222: You can fix that by adapting the limits so you're sure you're integrating over a smaller set to the right: $$\int_1^\infty \int_1^\infty \frac{dxdy}{x^2+y^2} \ge \int_3^\infty \int_{\pi/8}^{3\pi/8} \frac{rd\theta dr}{r^2} = \infty $$ – Troposphere Jun 23 '21 at 00:13

score 5 · Answer 2 · answered Jun 23 '21 at 08:03

5

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

answered Jun 23 '21 at 08:03

Greg Martin

78,820

(+1) As short as my approach – robjohn Jun 23 '21 at 13:11

score 2 · Answer 3 · answered Jun 22 '21 at 23:50

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 $\sum_{n=1}^\infty\frac{\pi}{2n^2}=\frac{\pi^3}{12}.$

The Double Basel Problem

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