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

∞∑n=110−n(n+1)/2=0.1010010001000010000010000001…

**Answer**

The number 0.1010010001000010000010000001… is transcendental.

Consider following three Jacobi theta series defined by

θ2(q)=2q1/4∑n≥0qn(n+1)=2q1/4∞∏n=1(1−q4n)(1+q2n)θ3(q)=∑n∈Zqn2=∞∏n=1(1−q2n)(1+q2n−1)2θ4(q)=θ3(−q)=∑n∈Z(−1)nqn2=∞∏n=1(1−q2n)(1−q2n−1)2

and for any m∈Z+, 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 i≠j. Let q∈C

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=110−n(n+1)/2=8√102θ2(1√10)−1

and using the fact 1√10 and 8√102 are both algebraic, we find the number at hand is transcendental.

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