3
$\begingroup$

I have a following series

$$ \sum\frac{1}{n^2+m^2} $$

As far as I understand it converges. I tried Cauchy criteria and it showed divergency, but i may be mistaken.

When I calculate it in matlab or Maxima it have a good behaviour and converge to finite number about 10.17615092535112.

The convergency plot is following:

enter image description here

  • 0
    Over which $n$ and $m$ do you take the sum? All $n$, $m$? Are they independent? I think then the series diverges...2012-05-24
  • 1
    Just to make sure (since I have no experience with matlab etc.) what does it give you for the series $\sum\frac 1n$?2012-05-24
  • 0
    thanks you, Simon, $\frac 1n$ also do not go to infinity and stay at really low values, like 12.09 at $10^5$ terms. That very strange.2012-05-24
  • 1
    That's what I expected. This has something to do with machine accuracy. Your computer can only represent numbers down to a certain number of digits. Hence for your computer the harmonic series is finite.2012-05-24
  • 1
    I bet you can even "show" that the harmonic series is constant after finitely many steps2012-05-24
  • 0
    I checked in matlab: $1/n$ diverges and 2D case $1/(n^2 + m^2)$ converges.2012-05-24
  • 0
    yep, this is something with machine accuracy, but nevertheless, those series have "logarithmic" divergency speed.2012-05-24
  • 0
    See also: [How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$](https://math.stackexchange.com/q/851302)2017-12-10

2 Answers 2

4

Just to give an answer. The series diverges since $$\frac{1}{n^2+m^2}\geq \frac{1}{(n+m)^2}$$ and therefore $$\sum_{n,m\in\mathbb N}\frac{1}{n^2+m^2}\geq \sum_{n,m\in\mathbb N}\frac{1}{(n+m)^2}.$$ For the RHS we have $$\sum_{n,m\in\mathbb N}\frac{1}{(n+m)^2}=\sum_k\sum_{n+m=k}\frac{1}{k^2}=\sum_k\frac{k-1}{k^2}\sim\sum_k\frac{1}{k}.$$

2

Matlabs sum for $\displaystyle \sum_{n=1}^{10^5} \frac{1}{n}$ is actually quite spot on, it's not a fault of machine accuracy. The series is called the Harmonic series and is known to diverge very slowly. In fact, it behaves approximately like $\gamma + \ln n$ where $\gamma$ is the Euler-Mascheroni constant, about $0.57721.$ At $n=10^5$ we indeed expect about $0.57721 + 5\ln 10 \approx 12.0901.$

Matlab is also not saying anything incorrect for the original series, its results were just interpreted incorrectly (in a similar way to the previous example). The sum grows quite slowly. Consider if we order the sum like this: $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2+n^2}.$$

By Residue Calculus methods, one can show that the inner sum is equal to $$ \frac{\pi m \coth(\pi m) -1}{2m^2}.$$

$\coth$ tends to $1$ exponentially quickly so one expects $$\sum_{m=1}^k \sum_{n=1}^{\infty} \frac{1}{m^2+n^2} \approx \frac{\pi \ln k}{2} + \frac{6\gamma \pi-\pi^2}{12}$$ for large $k.$