# Alternating sum of multiple zetas equals always 1?

This is more in the category of "recreational math"...
I was playing with multiple zetas, in the notation of $\zeta(k),\zeta(k,k),\zeta(k,k,k),\ldots$ as given in wikipedia.
Looking at the alternating sums $$A_m=\zeta(m)-\zeta(m,m)+\zeta(m,m,m)- \ldots + \ldots$$ it seems, that for any natural $m \ge 2$ we get $A_m = 1$. Even if we set $m=1$ and replace $\zeta(1)$ by the Euler-Mascheroni constant $\gamma$ (as done by Ramanujan in his summation) and compute the multiple zetas based on this we seem to approach $1$ when the number of multiple zetas increases without bound.

[update] It seems to be true for $m \gt 0$ and even for complex m, if $\Re(m) \gt 0$. [/update]

Q1: Is this true (and maybe even trivial)?
Q2: if the answer for Q1 is "true", then what is the range for m?

Example: the multiple zetas beginning at $\zeta(2)$ are $$\small \begin{array} {rr|l} \zeta(2)&=& 1.64493406685 \\ \zeta(2,2)&=& 0.811742425283 \\ \zeta(2,2,2)&=& 0.190751824122 \\ \zeta(2,2,2,2)&=& 0.0261478478177 \\ \zeta(2,2,2,2,2)&=& 0.00234608103546 \\ \ldots \\ \end{array}$$ The partial sums of the alternating series are $$\small \begin{array} {} 1.64493406685 \\ 0.833191641565 \\ 1.02394346569 \\ 0.997795617869 \\ 1.00014169890 \\ 0.999993270112 \\ 1.00000024599 \\ 0.999999992864 \\ 1.00000000017 \\ 0.999999999997 \\ 1.00000000000 \\ \ldots \end{array}$$ [update 2] The answer of @Achille Hui reminded me that it was of importance to remark that I compute the multiple zetas by the Newton-formula for the conversion between power-sums and elementary symmetric polynomials. Let $\operatorname{s2e}(v)$ be the function, which converts a vector v of powersums of consecutive exponents into a vector of elementary symmetric polynomials , then $$\operatorname{s2e}([\zeta(m),\zeta(2 m), \zeta(3 m),\ldots])=[\zeta(m),\zeta(m,m),\zeta(m,m,m),\ldots]$$ and the alternating sum $1-A_m$ can -at least formally- be rewritten as $$1-A_m = (1-1)(1-1/2^m)(1-1/3^m)\cdots$$ as @Achille hui has pointed out.

If that is a correct interpretation it is clear, that (if the product from the second term on converges for some m ) the complete result is zero for that same m.
This is backed by the observation, that if I remove the first (=zero)-parenthese, and then equivalently use $\zeta(\cdot)-1$ instead of $\zeta(\cdot)$ for the powersums, then for the well converging cases we seem to get again identity.

So if this is all "waterproof", then the question Q2 remains:

Q2: for which range of m (even from the complex numbers) do we have $1-A_m = 0$ ?

(I exclude so far all m , for which it can happen with some $k \in \mathbb N$ that $k\cdot m=1$) which simply means to avoid the occurence of $\zeta(1)$

• You know Gottfried, I really like reading your posts. You profess a "lack of formal education", and yet who can answer your weird and wonderful questions? – Douglas B. Staple May 28 '13 at 3:07
• @Douglas: That's a very nice compliment, thank you! Well - the formal education: true, and I confess that I've learnt so much from the discussions via the usenet/internet, the many advices and even from the pure reading of contributions of other people around. And by the great resource of online math material which makes it often much easier than the way through the lib... And in particular there are some such great (informal) advisors around here in MSE and MO - some of them I read like real friends... and am getting much thankful while you make me think about it. – Gottfried Helms May 28 '13 at 5:05

Given any sequence $(u_i)_{i\in\mathbb{Z}_{+}}$ with $u_i \in [0,1]$ and $\sum_{i=1}^{\infty} u_i < \infty$, we have:

$$\prod_{i=1}^{\infty} (1 - u_i ) = 1 + \sum_{k=1}^{\infty} (-1)^k \prod_{0 < i_1 < \ldots < i_k} u_{i_1}\cdots u_{i_k}$$

The RHS converges absolutely as a series because the $k^{th}$ term of it:

$$\prod_{0 < i_1 < \ldots < i_k} u_{i_1}\cdots u_{i_k} < \frac{1}{k!}\left(\sum_{i=1}^{\infty} u_i\right)^k$$

is bounded by the $k^{th}$ term of expansion of $\exp( \sum_{i=1}^{\infty} u_i)$.

Apply this to $u_n = n^{-m}$ where $m \ge 2$ and notice $u_1 = 1$, we get:

$$0 = 1 - \zeta(m) + \zeta(m,m) - \zeta(m,m,m) + \cdots$$

• Yes, that product-representation is what I just tried myself. It has the first factor as $(1-1)=0$ and thus the whole expression becomes zero and equivalently $1-A_m=0$ by my heuristics. This equivalence holds also, if I remove that zero-parenthese and on the other hand use $\zeta(...)-1$ instead of $\zeta(...)$ - at least for cases m where this converges with some dozen terms. So that product-representation would then introduce, that $1-A_m=0$ for all m. Hmm. But what about the divergent cases of negative m? – Gottfried Helms Apr 17 '13 at 6:02
• This type of argument only works for the case $\Re(m) > 1$ where you can express the zeta functions as a converging series. For $m < 0$, I don't even know whether $\zeta(\overbrace{m,\ldots,m}^{k\text{ terms}})$ converges to $0$ or not as $k \to \infty$. If yes, then some sort of arguments based on analytic continuation might work... – achille hui Apr 17 '13 at 6:55
• for one example of an m which produces divergent series see my new answer. – Gottfried Helms Apr 17 '13 at 7:35

We have $\zeta(k)=\sum_{m=1}^{\infty}\frac{1}{m^k}$

Define $\zeta(k,k)=\sum_{m=1}^{\infty}\frac{\prod_2 m}{m^k}$

where $\prod_2 m$ denotes the number of ways to express $m$ as a product of two distinct numbers(without counting orders, so $2\times 3= 3\times 2$ counted only once).

similarly define $\prod_r m$, the number of ways to express $m$ as a product of $r$ distinct numbers.

so that the $r$ th multiple zeta function value at $k$, satisfies $\zeta(k,\cdots,k) = \sum_{m=1}^{\infty} \frac{\prod_r m}{m^k}$.

Thus, the question boils down to checking whether $$1-\prod_2 m +\prod_3 m -+ \cdots + (-1)^{r-1} \prod_r m +\cdots = 0$$ if $m\geq 2$.

Considering that multiplying $1$ is also counted, we see that expression with $1$ and without $1$ cancels out. Thus our claim is true.

For example with $m=6$, we have

(-) for $1\times 6$, $2\times 3$

(+) for $1\times 2\times 3$. Here $2\times 3$ and $1\times 2\times 3$ cancels out.

For $m=12$,

(-) for $1\times 12$, $2\times 6$, $3\times 4$.

(+) for $1\times 2\times 6$, $1\times 3\times 4$. Here $2\times 6$ and $1\times 2\times 6$. $3\times 4$ with $1\times 3\times 4$ cancel out.

• Hi i707107 - thanks for the answer. I don't see the step, from the previous formula to where you say "it boils down...." . It might be a simple step, but I don't catch it right now. – Gottfried Helms Apr 17 '13 at 4:49
• That is just the calculation of Dirichlet series, coefficient of $1/m^s$. – Sungjin Kim Apr 17 '13 at 4:56

If found a partial answer in this paper by Ihara, Kaneko and Zagier. Formula (4.6) $$\exp_\star(\log_\circ(1+z)) =\frac{1}{1-z}$$ implies $1 = \zeta(m) - \zeta(m,m) + \zeta(m,m,m) - \ldots$ for all natural numbers $m \geq 2$. I'll explain the notation and how the formula implies the result.

Put $z_m := x^{m-1}y$, a word over the alphabet $\{x,y\}$ (or a monomial in non-commutating variables $x,y$ if you prefer). One defines $$\zeta(z_{m_1}\cdots z_{m_n}) := \zeta(m_1,\ldots,m_n)$$ for $m_i \geq 1$, $m_1 \geq 2$. One extends this to (finite or infinite) sums by linearity and puts $\zeta(1) = 1$. For example, $$\zeta(xxyxxxy + xy + 1) = \zeta(3,4) + \zeta(2) + 1.$$ One defines the harmonic product inductively by $$1 \star w = w \star 1 = w,\\ z_k w_1 \star z_l w_2 = z_k(w_1 \star z_l w_2) + z_l(z_k w_1 \star w_2) + z_{k+l}(w_1 \star w_2).$$ A theorem says that $\zeta$ is a homomorphism with respect to "$\star$", i.e. $$\zeta(w_1 \star w_2) = \zeta(w_1) \zeta(w_2).$$ One also defines the circle product by $$w_1 \circ w_2 = w_1 \star w_2 - w_1w_2 - w_2w_1,$$ so in particular $z_k \circ z_l = z_{k+l}$. Then, with $$\exp_\star(z) := 1 + z + \frac{1}{2} z\star z + \ldots,\\ \log_\circ(1+z) := z - \frac{1}{2} z \circ z + \frac{1}{3} z \circ z \circ z - \ldots,\\ \frac{1}{1-z} := 1 + z + z^2 + z^3 + \ldots$$ one has the formula stated above: $$\exp_\star(\log_\circ(1+z)) =\frac{1}{1+z}.$$ Putting $z := -z_m$ and applying $\zeta$, the right hand side gives $$\zeta(1 - z_m + z_m^2 - z_m^3 + \ldots) = 1 - \zeta(m) + \zeta(m,m) - \zeta(m,m,m) + \ldots$$ and the left hand side gives \begin{align*} \zeta(\exp_\star(\log_\circ(1-z_m))) &= \exp(\zeta(\log_\circ(1-z_m))) \\ &= \exp(-\sum_{n = 1}^\infty \frac{1}{n} \zeta(z_{mn}))\\ &= 0 \end{align*} since $\sum_{n = 1}^\infty \frac{1}{n} \zeta(z_{mn}) \geq \sum_{n = 1}^\infty \frac{1}{n} = \infty$. Equating the two yields $$1 - \zeta(m) + \zeta(m,m) - \zeta(m,m,m) + \ldots = 0.$$

• Wow, this is too much for the moment; I've downloaded the paper and shall go through it, but I doubt I'll get it completely. But lt's see. One question: does the above proof also allow to extend the sum-identity to non-integer arguments m of the zeta ? (Some empirical computations seem to indicate that the identity is still true for a wider range of m) – Gottfried Helms Apr 17 '13 at 5:01
• The proof doesn't generalize to non-integer arguments since $z_m = x^{m-1}y$ only makes sense if $m$ is a natural number. But maybe the result for natural $m$ implies the result for non-integer $m$ by the identity theorem. If $m \mapsto 1-A_m$ is holomorphic on a connected open subset $D$ of the Riemann sphere containing $\mathbb N$ and the point at infinity, then since $\mathbb N$ has the point at infinity as limit point, the identity theorem forces $1-A_m$ to vanish on all of $D$. – marlu Apr 17 '13 at 14:02
• Thank you very much again. I think I've got it now; that "star-" and "circle-" product is a nice tool. Can this method provide values for the negative (integer) $m = z_m$ as in the last couple of formulae? I find it also nice and interesting, that some intermediate steps give the same formula at which I arrived with my own manipulations (see the "update" at the end of my own answer) – Gottfried Helms Apr 19 '13 at 6:02

(This is not a new answer but just an additional example of data, reflecting the comment of @Achille hui concerning $\Re(m) \lt 1$)

Here is a list of the first few multiple zetas and the partial sums of the alternating series for $m=\sqrt{0.5}$ where the series-representation of $\zeta(m)$ is not convergent (and none of the powersums equals the $\zeta(1)$): $$\small \small \begin{array} {r|r|r} k & \zeta_k(m) \text{ : k'th mult.} \zeta() & \text{partial altern. sum} \\ \hline 1 & -2.85873148777 & -2.85873148777 \\ 2 & 2.57580401985 & -5.43453550762 \\ 3 & 0.938727829684 & -4.49580767793 \\ 4 & -4.02972075888 & -0.466086919058 \\ 5 & 3.46585761472 & 2.99977069567 \\ 6 & -0.626085735043 & 3.62585643071 \\ 7 & -1.29780958922 & 2.32804684149 \\ 8 & 1.28297262963 & 1.04507421187 \\ 9 & -0.436367742286 & 0.608706469581 \\ 10 & -0.122662257866 & 0.731368727447 \\ 11 & 0.200808483827 & 0.932177211274 \\ 12 & -0.0882083797854 & 1.02038559106 \\ 13 & 0.00437308895501 & 1.02475868001 \\ 14 & 0.0155665983724 & 1.00919208164 \\ 15 & -0.00874359614267 & 1.00044848550 \\ \cdots & \cdots &\cdots \\ 60 & -1.92516635703E-31 & 1.00000000000 \\ 61 & -2.39632638182E-33 & 1.00000000000 \\ 62 & 5.31122663432E-33 & 1.00000000000 \\ 63 & -9.45648541011E-34 & 1.00000000000 \\ 64 & 4.89013667885E-35 & 1.00000000000 \\ \cdots & \cdots &\cdots \\ \end{array}$$ This looks similar with any value $m \gt 0$, only that the index k from where the multiple zetas begin to vanish may become higher. We can even use complex m, for instance from the nontrivial roots and can directly or Eulersum the alternating series ...

Here is a list using the negative value $m=-1$. The alternating series needs now a procedure for divergent summation, like Borel-summation (it seems, that Euler-summation may not suffice) . We get $$\small \small \begin{array} {r|r|r} k & \zeta_k(-1) \text{ : k'th mult.} \zeta() & \text{partial altern. sum} \\ & & \text{divergent summation } \\ \hline 1 & -0.0833333333333 & -0.0833333333333 \\ 2 & 0.00347222222222 & -0.0848429951691 \\ 3 & 0.00268132716049 & -0.0852728890865 \\ 4 & -0.000229472093621 & -0.0853089197972 \\ 5 & -0.000784039221720 & -0.0852036275208 \\ 6 & 0.0000697281375837 & -0.0850590273088 \\ 7 & 0.000592166437354 & -0.0849170033670 \\ 8 & -0.0000517179090826 & -0.0847934738378 \\ 9 & -0.000839498720672 & -0.0846927757523 \\ 10 & 0.0000720489541602 & -0.0846141577213 \\ 11 & 0.00191443849857 & -0.0845548301287 \\ 12 & -0.000162516262784 & -0.0845114155482 \\ 13 & -0.00640336283381 & -0.0844806229031 \\ 14 & 0.000540164767893 & -0.0844595354062 \\ 15 & 0.0295278809457 & -0.0844457052452 \\ 16 & -0.00248174360026 & -0.0844371535103 \\ 17 & -0.179540117061 & -0.0844323265957 \\ 18 & 0.0150561130400 & -0.0844300358670 \\ 19 & 1.39180109327 & -0.0844293944146 \\ 20 & -0.116546276599 & -0.0844297577262 \\ 21 & -13.3979854551 & -0.0844306713024 \\ 22 & 1.12080446429 & -0.0844318261930 \\ 23 & 156.801412704 & -0.0844330223409 \\ 24 & -13.1078630226 & -0.0844341390840 \\ 25 & -2192.55553609 & -0.0844351119332 \\ \cdots & \cdots &\cdots \\ 60 & -5.03959577267E29 & -0.0844375513246 \\ 61 & -5.42046040306E32 & -0.0844375514123 \\ 62 & 4.51885202016E31 & -0.0844375514570 \\ 63 & 5.19276945836E34 & -0.0844375514751 \\ 64 & -4.32892308840E33 & -0.0844375514782 \\ \cdots & \cdots &\cdots \\ \end{array}$$
Using Wolframalpha for a guess for the result we find that possibly $$A_{-1} \underset{\text{guessed}}{=} 1-{\exp(1) \over \sqrt{2\pi}} \sim -0.084437551419227546612$$

[update 2] It seems, there is an alternative formula for the computation of the $A_m$ possible, however I've still no well working summation procedure for the resulting alternative, but still divergent, series.

First, for smoothing everything let's introduce the function $$B(m) = 1- A_m = 1- \zeta(m) + \zeta(m,m) - \zeta(m,m,m) + \ldots - \ldots$$ Next, let's formally rewrite this as infinite product $$P(m) = (1-1^{m})(1-1/2^{m}) (1-1/3^{m}) \ldots \underset{\text{formally}}{=}B(m)$$ Then the log of $P(m)$ is a sum of logs $$L(m)=\log(P(m)) = \log(1-1^{m}) + \log(1-1/2^{m}) + \log(1-1/3^{m}) \ldots$$which can be rewritten as a double series involving the well known Mercator-series for the logarithm. Then we change order of summation; this seem to result -again formally- in the following sum of zetas $$L(m)= -\zeta(1m)/1 - \zeta(2m)/2 - \zeta(3m)/3 - \ldots$$ Of course, all this is just handwaving and needs formal justification.

However, just feeding the latter formula, using $m=-1$ into a summation-procedure for divergent summation I arrive at the same result as with the original computation: $$B(-1) = \exp(L(-1)) \underset{\mathfrak N}{=} {\exp(1) \over \sqrt{2 \pi}}$$ where $\mathfrak N$ indicates a customized Noerlund procedure for divergent summation.

So we may take the hypothesis as granted, that this replacement of the original computation is valid. Unfortunately it is still strongly diverging, and the rate of growth increases with the absolute value of m (unless of course m is a multiple of $-2$). Here are a couple more rough estimates: $$\small \begin{array} {r|lr} m & B(m) (\text{ using } \mathfrak N ) \\ \hline -1 & 1.0844375514192275466 & \qquad (= 1-A_{-1} )\\ -1/2 & 1.2904007518681174634 \\ -1/3 & 1.48044921903 \\ -1/4 & 1.65184851943 \end{array}$$

• (+1) interesting. $A_{-1}$ meta converges but to a different number. It will be interesting to see the values of other $A_{-m}$ and whether there is any pattern... – achille hui Apr 17 '13 at 9:20
• @achillehui: Yes, that is interesting! Unfortunately my summation-procedure does not well approximate final values, so my heuristical data are much spurious. For $m \lt -1$ the divergence is pretty hard (except for the trivial $m=-2n$ where we have zeros only). At the moment I need a better tool for that summations... – Gottfried Helms Apr 17 '13 at 10:56

At the following MSE link on MZVs and the PET an algorithm was used to compute MZVs that is not as good as what has been presented in this thread, but more importantly, it was proved that $$\zeta(s,s,\ldots,s) = Z(P_n)(\zeta(s);\zeta(2s);\ldots;\zeta(ns))$$ where $Z(P_n)$ is the cycle index $Z(A_n)-Z(S_n)$ of the unlabelled set operator $\mathfrak{P}_{=n}$ which has $Z(P_0) = 1$ and obeys the recurrence $$Z(P_n) = \frac{1}{n}\sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l})$$ which can be proved by direct enumeration according to the cycle that contains $n.$ This operator is standard in applications of the Polya Enumeration Theorem and can be used to prove the conjecture from above.

Re-stating the conjecture in terms of the PET we obtain the equivalent form which says that $$\sum_{n\ge 0} (-1)^n Z(P_n)(\zeta(s);\zeta(2s);\ldots;\zeta(ns)) = 0.$$

To prove this we introduce (this is a standard technique) the generating function $$F(y) = \sum_{n\ge 0} (-1)^n y^n Z(P_n)(\zeta(s);\zeta(2s);\ldots;\zeta(ns))$$ so that we are interested in $F(1).$ We will use the simplified notation $$Z(P_n)(\zeta(s))$$ for the substituted cycle index according to PET rules.

Differentiate to obtain $$F'(y) = \sum_{n\ge 1} (-1)^n n y^{n-1} Z(P_n)(\zeta(s)).$$ Apply the recurrence from above to get $$\sum_{n\ge 1} (-1)^n y^{n-1} \sum_{l=1}^n (-1)^{l-1} \zeta(ls) Z(P_{n-l})(\zeta(s)).$$ This is $$\sum_{n\ge 1} (-1)^n y^{n-1} \sum_{l=1}^n (-1)^{l-1} \zeta(ls) (-1)^{n-l} [y^{n-l}] F(y) \\ = \sum_{l\ge 1} (-1)^{l-1} \zeta(ls) \sum_{n\ge l} (-1)^n y^{n-1} (-1)^{n-l} [y^{n-l}] F(y) \\ = \sum_{l\ge 1} (-1)^{l-1} \zeta(ls) y^{l-1} \sum_{n\ge l} (-1)^n (-1)^{n-l} y^{n-l} [y^{n-l}] F(y) \\ = - \sum_{l\ge 1} \zeta(ls) y^{l-1} \sum_{n\ge l} y^{n-l} [y^{n-l}] F(y) \\= \left(- \sum_{l\ge 1} \zeta(ls) y^{l-1} \right) F(y).$$

Formally integrating the equation $$(\log F(y))' = - \sum_{l\ge 1} \zeta(ls) y^{l-1}$$ we thus obtain $$\log F(y) = - \sum_{l\ge 1} \zeta(ls) \frac{y^l}{l}$$ where we have used the fact that $\log F(0) = \log Z(P_0) = \log 1 = 0.$

When $s>1$ the unsigned sum term converges in $[-1,1)$ and at the upper end $$\lim_{y\rightarrow 1-} \sum_{l\ge 1} \zeta(ls) \frac{y^l}{l} = +\infty$$ where the harmonic series was used to show divergence to $+\infty$ at $y=1.$

We finally have $$F(y) = \exp\left(- \sum_{l\ge 1} \zeta(ls) \frac{y^l}{l}\right)$$ and it follows that in the limit $$F(1) = \exp(-\infty) = 0$$ as was to be shown.

This proof is equivalent to what has been presented by the other contributors but I wanted to make sure the relation to the Polya Enumeration Theorem be made prominent. As was pointed out in the thread referenced at the beginning this computation goes through with an ordinary generating function instead of a Dirichlet series, it is, I quote, pure symmetry.

• With an OGF $A(z)$ instead of a Dirichlet series we would have $$F(y) = \exp\left(-\sum_{l\ge 1} A(z^l) \frac{y^l}{l}\right).$$ – Marko Riedel May 6 '14 at 1:14