14
$\begingroup$

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?

$\endgroup$
4
  • 5
    $\begingroup$ Of course, it is identity 67 on mathworld.wolfram.com/RiemannZetaFunction.html $\endgroup$ Feb 11, 2011 at 17:02
  • 12
    $\begingroup$ @Dror: why of course? $\endgroup$ 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? $\endgroup$ 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). $\endgroup$ Feb 16, 2011 at 10:14

1 Answer 1

26
$\begingroup$

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.

$\endgroup$
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. $\endgroup$ Feb 11, 2011 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.