Erdős Problem #250
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The number \[ S=\sum_{n=1}^{\infty}\frac{\sigma(n)}{2^n}, \qquad \sigma(n)=\sum_{d\mid n}d, \] is not merely irrational: it is transcendental. We give the complete deduction from Nesterenko's theorem on Eisenstein series. That deep algebraic-independence theorem is an external input and is not reproved here.
Restatement from Math-5530-A1 r1, with the source manuscript’s qualifications.
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GPT 6 Astra Ultra attempt on Math-5530
Erdős Problem 250: irrationality of a divisor-sum series
1) Statement and scope
The number \[ S=\sum_{n=1}^{\infty}\frac{\sigma(n)}{2^n}, \qquad \sigma(n)=\sum_{d\mid n}d, \] is not merely irrational: it is transcendental. We give the complete deduction from Nesterenko's theorem on Eisenstein series. That deep algebraic-independence theorem is an external input and is not reproved here.2) Convergence and the modular-series identity
For $r\geq1$, let $\sigma_r(n)=\sum_{d\mid n}d^r$. The bound $\sigma_r(n)\leq n^{r+1}$ ensures convergence of the following series for $|q|<1$: \[ \begin{split} P(q)&=1-24\sum_{n\geq1}\sigma_1(n)q^n,\\ Q(q)&=1+240\sum_{n\geq1}\sigma_3(n)q^n,\\ R(q)&=1-504\sum_{n\geq1}\sigma_5(n)q^n. \end{split} \] In particular, \[ P(1/2)=1-24S. \] The nonnegative divisor expansion also gives the Lambert-series description \[ S=\sum_{d\geq1}\sum_{j\geq1}\frac{d}{2^{dj}} =\sum_{d\geq1}\frac{d}{2^d-1}. \] Both the interchange and the identity follow from absolute convergence (or directly from Tonelli's theorem).3) Applying the external theorem
Nesterenko's theorem states that for each complex $q$ with $0<|q|<1$, \[ \operatorname{trdeg}_{\mathbb Q} \mathbb Q\left(q,P(q),Q(q),R(q)\right)\geq3. \] At $q=1/2$, adjoining $q$ adds no transcendence degree. Thus the three numbers $P(1/2),Q(1/2),R(1/2)$ are algebraically independent over $\mathbb Q$. In particular $P(1/2)$ is transcendental: otherwise the field generated by all three would have transcendence degree at most two. If $S$ were algebraic, then $1-24S=P(1/2)$ would be algebraic, a contradiction. This proves the asserted transcendence.4) Scope and sources
Numerical approximation does not establish this conclusion. For example the entirely elementary tail bound \[ 0<S-\sum_{n=1}^M\frac{\sigma(n)}{2^n} \leq\sum_{n>M}\frac{n^2}{2^n} =\frac{M^2+4M+6}{2^M} \] certifies convergence but not irrationality. The transcendence step above depends essentially on the stated theorem.Sources: https://www.erdosproblems.com/250; Yu. V. Nesterenko, Modular functions and transcendence questions, Sbornik Math. 187 $1996$, 1319--1348, https://doi.org/10.1070/SM1996v187n09ABEH000158; W. Zudilin, Diophantine problems for $q$-zeta values, https://arxiv.org/abs/math/0206179, which also records this transcendence consequence for the divisor-sum series. This is an exposition of the known resolution, not a new algebraic-independence proof or a local formal verification. The full question follows from the established external theorem.