Is 0.1010010001000010000010000001…0.1010010001000010000010000001 \ldots transcendental?

Does anyone know if this number is algebraic or transcendental, and why?

n=110n(n+1)/2=0.1010010001000010000010000001

Answer

The number 0.1010010001000010000010000001 is transcendental.

Consider following three Jacobi theta series defined by
θ2(q)=2q1/4n0qn(n+1)=2q1/4n=1(1q4n)(1+q2n)θ3(q)=nZqn2=n=1(1q2n)(1+q2n1)2θ4(q)=θ3(q)=nZ(1)nqn2=n=1(1q2n)(1q2n1)2
and for any mZ+, k{2,3,4}, use
Dmθk(q) as a shorthand for
(qddq)mθk(q).

Based on Corollary 52 of a survey article Elliptic functions and Transcendence by M. Waldschmidt in 2006,

Let i,j and k{2,3,4} with ij. Let qC
satisfy 0<|q|<1. Then each of the two fields
Q(q,θi(q),θj(q),Dθk(q)) and Q(q,θk(q),Dθk(q),D2θk(q))
has transcendence degree 3 over Q

We know for any non-zero algebraic q with |q|<1, the three θk(q), in particular θ2(q) is transcendental. Since

n=110n(n+1)/2=8102θ2(110)1

and using the fact 110 and 8102 are both algebraic, we find the number at hand is transcendental.

Attribution
Source : Link , Question Author : Raffaele , Answer Author : Grigorios Kostakos

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