# Are any values of this sum involving the Thue Morse sequence known?

Let $$t_n$$ denote the $$n^{\rm th}$$ term in the Thue Morse sequence. Note that $$t_n=1$$ if the number of $$1$$s in the binary expansion of $$n$$ is odd, $$0$$ otherwise. Now define a variant of the Riemann Zeta function as follows:

$$\zeta_{TM}(s) = \sum_{n\geq0} \frac{t_{n}}{(n+1)^s}$$ for $$\rm{Re}(s)>1$$. Are any values of $$\zeta_{TM}(s)$$ known? Is there some kind of closed form formula for this (I highly doubt there is, but one never knows).

A closely related sum for which a value is known is:

$$\sum_{n\geq1} \frac{s_{n}}{n(n+1)} = 2\ln2,$$

where $$s_n$$ is the binary sum-of-digits function. Another related sum gives the Prouhet-Thue-Morse constant, which has been shown to be transcendental:

$$\tau =\sum _{{n\geq0}}{\frac {t_{n}}{2^{{n+1}}}}=0.412454033640\ldots$$

• "but one never knows" you can disprove the existence of closed form formulas – mathworker21 May 14 at 9:27
• Sure. What I meant by that was an informal "I don't know if such a formula exists in the vast ocean of published literature". – Klangen May 14 at 9:31
• Related post on Mathematica.Stackexchange: mathematica.stackexchange.com/questions/198604/… – Klangen May 18 at 8:49

There's probably no "closed form" for any $$s>1$$. Curiously, though, the sum can be evaluated for integers $$s \leq 0$$, in the following sense. The function $$\zeta_{TM}$$ extends to an analytic function on $${\bf C} \backslash \{ 1 \}$$, with a simple pole at $$s=1$$ of residue $$1/2$$, and taking rational values at integers $$s \leq 0$$, starting $$\zeta_{TM}(0) = -1/4$$, $$\zeta_{TM}(-1) = -1/24$$, $$\zeta_{TM}(-2) = 0$$, $$\zeta_{TM}(-3) = +1/240$$, and in general $$\zeta_{TM}(s) = \zeta(s) / 2$$ for integers $$s \leq 0$$ (so in particular $$\zeta_{TM}$$ inherits the "trivial zeros" of $$\zeta$$ at $$s = -2, -4, -6, \ldots$$).
It is more convenient to work with the Dirichlet series whose $$(n+1)^{-s}$$ coefficient is $$1 - 2 t_n = (-1)^{t_n}$$, because the generating function for $$(-1)^{t_n}$$, call it $$T(z) = \sum_{n=0}^\infty (-1)^{t_n} z^n,$$ factors as an infinite product: $$T(z) = (1-z) (1-z^2) (1-z^4) (1-z^8) \cdots = \prod_{m=0}^\infty \bigl( 1 - z^{2^m} \bigr).$$ So define $$Z_{TM}(s) = \zeta(s) - 2 \zeta_{TM}(s) = \sum_{n=0}^\infty \frac{(-1)^{t_n}}{(n+1)^s}.$$ The usual Mellin-transform trick gives an integral formula: $$\Gamma(s) Z_{TM}(s) = \sum_{n=0}^\infty (-1)^{t_n} \! \int_0^\infty x^{s-1} e^{-(n+1)x} \, dx = \int_0^\infty x^{s-1} e^{-x} T(e^{-x}) \, dx.$$ This gives an analytic continuation of $$\Gamma(s) Z_{TM}(s)$$ to the entire complex plane, because $$T(e^{-x})$$ decays faster than any power of $$x$$ as $$x \to 0$$: each factor $$1 - e^{-2^m x}$$ of the infinite product is $$O_m(x)$$ and in $$(0,1)$$. Since $$\Gamma(s)$$ has no zeros, but does have simple poles at $$s = 0, -1, -2, -3, \ldots$$, it follows that $$Z_{TM}$$ is an entire function with simple zeros at the same $$s$$, and no other real zeros (the integral for $$\Gamma(s) Z_{TM}(s)$$ is plainly positive for all real $$s$$). Since $$Z_{TM} = \zeta - 2 \zeta_{TM}$$, we conclude that $$\zeta_{TM}(s) = \frac12 \zeta(s)$$ at those $$s$$, as claimed.
[The integral formula can also be used to compute $$Z_{TM}(s)$$, and thus also $$\zeta_{TM}(s)$$, to high precision; for example, using gp's "intnum" function we find $$Z_{TM}(2) = 0.6931534522\ldots$$ (this is not $$\log 2$$, though it's quite close -- the difference is $$\lt 10^{-5}$$), so $$\zeta_{TM}(2) = (\zeta(2) - Z_{TM}(2)) / 2 = 0.4758903073\ldots$$.]