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I'm trying to compute the sum of this infinite series, but I can't really find a common ratio for it. I've tried breaking it up into the sum of three separate series but to no avail. I'm not sure that I understand exactly how the series works. I need to put it into a form that has $\pi$, but I only lead on that was with the Riemann zeta function. Here's the sum.

$$\sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\sum_{z=1}^{\infty}\frac{100z^2yx}{z^3y^4x^3+z^3y^3x^4+z^4y^3x^3}$$

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2 Answers 2

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Forget the 100 for a bit.

Clearly you will get the same answer if you permute $x,y,z$. So the sum is one-third of the same sum with $x^2 yz +y^2 zx +z^2 xy$ on top.

Then the term we are summing reduces at once to $\frac{1}{x^2}\frac{1}{y^2}\frac{1}{z^2}$.

So we'll get something like $\frac{100}{3}(\frac{\pi^2}{6})^3$.

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    $\begingroup$ The denominator is symmetric and is $(xyz)^3 (x+y+z)$. The numerator is $z^2 xy$; these are just names and we'd get the same result for numerator $x^2 yz$ and for numerator $y^2 zx$; so add the three together (to get 3 times the sum), then the numerator is $xyz(x+y+z)$, and everything simplifies. $\endgroup$ Nov 3, 2019 at 17:55
  • $\begingroup$ That's incredible, and so much simpler than I thought. Does the process of interchanging the variables in the numerator have a specific name? How did you know to do that? Also, is the sum of 1/x^2 the Fourier series? So if you have a triple sum of the Fourier series with all the same terms you can call it $(\frac{\pi}{6})^3$? $\endgroup$
    – James R
    Nov 3, 2019 at 18:11
  • $\begingroup$ Wow, that's smart! +1 $\endgroup$
    – amsmath
    Nov 4, 2019 at 1:28
  • $\begingroup$ @JamesR, I'd just say I symmetrised the sum, but have no idea if that's the correct name. Once the summand is a product of $f(x), g(y), h(z)$ the triple sum becomes a product of three sums formally; and as the terms are all $O(\frac{1}{n^2}$ everything is sufficiently convergent. Fourier theory indeed gives the vlaue of the sum, but there must b e other ways too. $\endgroup$ Nov 4, 2019 at 7:29
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This is only a partial answer with which you might want to proceed. Let me denote your sum by $S$. Then \begin{align} S &= \sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\sum_{z=1}^{\infty}\frac 1{zy^2x^2}\cdot\frac{100}{x+y+z}\\ &= 100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\frac 1{y^2}\sum_{z=1}^{\infty}\frac 1{z}\cdot\frac{1}{x+y+z}\\ &= 100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\frac 1{y^2}\frac{1}{x+y}\sum_{z=1}^{\infty}\left(\frac{1}{z}-\frac{1}{x+y+z}\right)\\ &=100\sum_{x=1}^{\infty}\frac 1{x^2}\sum_{y=1}^{\infty}\left(\frac{1}{x+y}\sum_{z=1}^{x+y}\frac{1}{z}\right)\frac 1{y^2}\\ &=100\sum_{x=1}^{\infty}\sum_{y=1}^{\infty}\frac{M(x+y)}{(xy)^2}, \end{align} where $M(n) = \frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}$.

EDIT: I just saw that setting $p=x+y$ you get $$ S = \sum_{p=2}^\infty M(p)\sum_{x=1}^{p-1}\frac 1{x^2(p-x)^2}. $$ Maybe you can do the partial fraction decomposition on the inner series again to simplify it... I have to go now. Good luck!

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  • $\begingroup$ Thank you so much! I'll try that approach. $\endgroup$
    – James R
    Nov 3, 2019 at 17:10

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