15
$\begingroup$

Does: $$\sum_{1 \leq i<j} \frac{1}{i j^2} = \sum_{1 \leq k} \frac{1}{k^3}?$$

Motivation: Call the above sum $S$, and let $$T := \sum_{ GCD(i,j)=1} \frac{1}{\max(i,j) i j}.$$ The sum $T$ came up in a computation on Jim Propp's question here. Numerical computation suggested that $T$ is extremely close to $3$.

It is not hard to show that $$T = \zeta(3)^{-1} \sum \frac{1}{\max(i,j) i j} = \zeta(3)^{-1} \left( \sum_{k} \frac{1}{k^3} + 2 \sum_{i<j} \frac{1}{i j^2} \right) = 1 + 2 \zeta(3)^{-1} S,$$ by breaking into cases according to whether $i<j$, $i=j$ or $i>j$. So $T=3$ iff $S=\zeta(3)$.

As I describe in the above linked thread, numerical computations suggest that the sums agree to $20$ digits of accuracy. What is going on?

Improve this question
$\endgroup$
6
  • 5
    $\begingroup$ Of course, it is identity 67 on mathworld.wolfram.com/RiemannZetaFunction.html $\endgroup$
    – Dror Speiser
    Commented Feb 11, 2011 at 17:02
  • 13
    $\begingroup$ @Dror: why of course? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 11, 2011 at 17:57
  • 1
    $\begingroup$ This question appears to be related to a recent one of mine: mathoverflow.net/questions/50253. I'm wondering if one of the solutions given there can be used to give another perspective on this result? $\endgroup$
    – Mike Spivey
    Commented Feb 11, 2011 at 19:26
  • 2
    $\begingroup$ @Mariano: aside from the fact that I remembered seeing the identity on that page a few years back, finding it again was routine: google "Riemann Zeta Function", second hit, skimming to about mid page, and there, but not over - first reference Stark, jstor, and he references Klamkin, as did David below. As can be seen from the timestamps, this was done in under 5 minutes (given that I remembered seeing it). $\endgroup$
    – Dror Speiser
    Commented Feb 16, 2011 at 10:14
  • $\begingroup$ See also the answers here: mathoverflow.net/questions/458221 $\endgroup$
    – GH from MO
    Commented Nov 12, 2023 at 7:46

1 Answer 1

Reset to default
27
$\begingroup$

Hi David,

This is the first example of a multiple zeta identity. Your sum S is just $\zeta(1,2)$, where the multiple zeta value is defined by: $$\zeta(s_1, s_2, \ldots, s_k) = \sum_{0 < n_1 < n_2 < \cdots n_k} \left( \prod_{i=1}^k n_i^{-s_i} \right).$$

Your identity $\zeta(1,2) = \zeta(3)$ was discovered by Euler according to Wikipedia.

Improve this answer
$\endgroup$
1
  • 5
    $\begingroup$ Thanks Marty! For those with JSTOR access, a very clean proof is given at jstor.org/stable/2308345 . For those with access to a good library, this is The American Mathematical Monthly, Vol. 59, No. 7, Aug. - Sep., 1952, p. 471 , problem proposed by M. Klamkin, solution by R. Steinberg. $\endgroup$
    – David E Speyer
    Commented Feb 11, 2011 at 17:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .