I What's the name of this series?

[puzzled fish]

I found this series, when I tried to evaluate the net Newtonian gravitational force on a mass at rest upon one vertex of a cube while all the other masses were arranged on an orthogonal lattice inside the cube:
$\sum\limits_{k=1}^{\infty} \sum\limits_{j=0}^{\infty} \sum\limits_{i=0}^{\infty} \frac 1 {i^2 + j^2 + k^2}$
I tried to evaluate, $\sum\limits_{j=1}^{\infty} \sum\limits_{i=0}^{\infty} \frac 1 {i^2 + j^2}$ with a quick VB macro ( Mathematica was very slow ) up to i = j = 500000 and it seems to converge. The amount of calculations for the triple series is enormous, I haven't done it, but it still converges I think.
Sorry, I don't know how to fix this Latex code.

 

[mfb]

You can include LaTeX by putting ## or $[/color]$ around formulas, done in the quote here:

I found this series, when I tried to evaluate the net Newtonian gravitational force on a mass at rest upon one vertex of a cube while all the other masses were arranged on an orthogonal lattice inside the cube.
$$\sum\limits_{k=0}^{\infty} \sum\limits_{j=0}^{\infty} \sum\limits_{i=0}^{\infty} \frac 1 {i^2 + j^2 + k^2}$$ with the first term i = j = k = 0 removed.
I tried to evaluate, $$\sum\limits_{j=0}^{\infty} \sum\limits_{i=0}^{\infty} \frac 1 {i^2 + j^2}$$ ( first term removed ) with a quick VB macro ( Mathematica was very slow ) up to i = j = 500000 and it seems to converge. The amount of calculations for the triple series is enormous, I haven't done it, but it still converges I think.

The individual sums have analytic expressions: WolframAlpha result. You can see if you can sum over these expressions again.

 

[Orodruin]

It does not converge.

 

[mfb]

Oh, good point.

It is similar to an integral over $\frac{1}{r^2}$ in 3-dimensional space.

 

[puzzled fish]

So, it does converge...

 

[Orodruin]

So, it does converge...

No. And by forgetting that the forces come with different directions you made it even less convergent.

 

[jostpuur]

For a moment I tried to get a rigorous lower bound for the two dimensional sum by starting from

<br /> \frac{1}{x^2 + y^2} \leq \frac{1}{\lfloor x\rfloor^2 + \lfloor y\rfloor^2}<br />

but unfortunately a bunch of technical difficulties arise from the polar coordinates working nicely around the origin (0,0), and the original sum having instead the point (1,1) with a special meaning, and eventually I gave up thinking that this isn't such an interesting problem. Anyway, it looks like that if you invest enough effort to this, you can come up with some rigorous lower bound eventually. If puzzled fish is still taking this sum seriously, perhaps he should take finding that lower bound as challenge?

Meanwhile, I wrote down the following crude approximation:

<br /> \sum_{n,m=1}^{N-1} \frac{1}{n^2 + m^2} \approx \int\limits_1^N dx\int\limits_1^N dy\frac{1}{x^2 + y^2}<br /> \approx \int\limits_0^{\frac{\pi}{2}}d\theta \int\limits_1^N dr \;r \frac{1}{r^2} = \frac{\pi}{2}\ln(N)<br />

puzzled fish, does it look like that your numerical computation are in agreement with this approximation, at least concerning the magnitudes?

 

[puzzled fish]

For a moment I tried to get a rigorous lower bound for the two dimensional sum by starting from

<br /> \frac{1}{x^2 + y^2} \leq \frac{1}{\lfloor x\rfloor^2 + \lfloor y\rfloor^2}<br />

but unfortunately a bunch of technical difficulties arise from the polar coordinates working nicely around the origin (0,0), and the original sum having instead the point (1,1) with a special meaning, and eventually I gave up thinking that this isn't such an interesting problem. Anyway, it looks like that if you invest enough effort to this, you can come up with some rigorous lower bound eventually. If puzzled fish is still taking this sum seriously, perhaps he should take finding that lower bound as challenge?

Meanwhile, I wrote down the following crude approximation:

<br /> \sum_{n,m=1}^{N-1} \frac{1}{n^2 + m^2} \approx \int\limits_1^N dx\int\limits_1^N dy\frac{1}{x^2 + y^2}<br /> \approx \int\limits_0^{\frac{\pi}{2}}d\theta \int\limits_1^N dr \;r \frac{1}{r^2} = \frac{\pi}{2}\ln(N)<br />

puzzled fish, does it look like that your numerical computation are in agreement with this approximation, at least concerning the magnitudes?

Thank you for your answer. Very foolish question of me indeed! Next time I shall remember to take the integral rest sooner..