Is it reasonable to assert that $0^k - 1^k - 2^k + 3^k - 4^k + 5^k + 6^k - 7^k - ... = 0$ for all $k > 1$?

Here the signs are given by the Thue-Morse sequence; that is, the sign of $m^k$ is $+$ or $-$ according to whether the number of 1's in the binary expansion of $m$ is even or odd.

I'm aware that the theory of divergent series is actually an assortment of theories that don't always agree (even though they usually do), so that it isn't necessarily meaningful to ask what "the" value of a divergent series is without specifying what sort of regularization one has in mind. Still, the following reasoning strikes me as having some evidentiary force.

Take $f(q):=(1-q)(1-q^2)(1-q^4)(1-q^8)\cdots$ for $0 \leq q \leq 1$. Because of the numerous factors of $1-q$, we have $f^{(k)}(1) = 0$ for all $k \geq 0$. (Fussy technicality: $f^{(k)}(1)$ needs to be defined as a one-sided derivative because $f(q)$ is undefined for $q>1$.) If we repeatedly differentiate $f(q)$ term-by-term, and then plug in $q=1$, we get divergent sums of the form $p(0) - p(1) - p(2) + p(3) - p(4) + p(5) + p(6) - p(7) - ...$ where $p(\cdot)$ is a polynomial; since $f^{(k)}(1) = 0$, we are "justified" in asserting that the divergent sum vanishes. Since these polynomials form a basis for the space of all polynomials, we have $p(0) - p(1) - p(2) + p(3) - p(4) + p(5) + p(6) - p(7) - ... = 0$ for all polynomials $p(x)$, and hence in particular for $p(x) = x^k$.

Compare with the fairly well-known fact (source?) that if $n=2^d$ then $0^k - 1^k - 2^k + 3^k - ... \pm (n-1)^k$ vanishes for all $k$ between 0 and $d-1$.

  • $\begingroup$ In light of your last paragraph, at least a subsequence of partial sums are 0. $\endgroup$ – Fan Zheng Jan 9 '17 at 3:10
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    $\begingroup$ "If we repeatedly differentiate f(q) term-by-term, and then plug in q=1, we get divergent sums of the form p(0)−p(1)−p(2)+p(3)−p(4)+p(5)+p(6)−p(7)−...p(0)−p(1)−p(2)+p(3)−p(4)+p(5)+p(6)−p(7)−... ” what does this means? can you make it more clear ?the computing process is confusing for me. $\endgroup$ – D.SUPPER Jan 9 '17 at 3:29
  • $\begingroup$ Key word: Abel summation. (I just realized that there is more than one thing in math called Abel summation.) $\endgroup$ – Fan Zheng Jan 9 '17 at 3:36
  • 1
    $\begingroup$ Source = Prouhet, I believe. The French call the Morse-Thue sequence the Prouhet-Morse-Thue sequence (Prouhet = 1855 or thereabouts). $\endgroup$ – Anthony Quas Jan 9 '17 at 6:27
  • $\begingroup$ If we differentiate $1 - q - q^2 + q^3 - q^4 + \dots$ with respect to $q$, we get $0-1-2q+3q^2-4q^3+\dots$; plugging in $q=1$ we get the divergent series $0-1-2+3-4+\dots$, which is $p(0)−p(1)−p(2)+p(3)−p(4)+...$ with $p(x)=x$. If instead we differentiate $1 - q - q^2 + q^3 - q^4 + \dots$ twice with respect to $q$ before plugging in $q=1$, we get $0-0-2+6q-12q^2+\dots$, which is $p(0)−p(1)−p(2)+p(3)−p(4)+...$ with $p(x)=x(x-1)$. $\endgroup$ – James Propp Jan 10 '17 at 4:06

As Fan Zheng points out in his comment, if the divergent series is interpreted in the sense of Abel summation then the value is 0 (and the argument given in the fourth paragraph of the question contains most of the ideas required for a rigorous proof). So the assertion is indeed reasonable.


You might be interested in the answer/solution to my similar question Cesaro(?)/Euler(?) - summation of the $s(p)=\sum_{k=0}^\infty (-1)^{H(k)} (1+k)^p$ for $p=1,2,3,...$ (where $H(k)$ is the Hamming-weight) given by Lucia https://mathoverflow.net/a/168795/7710


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