This page was last edited 28 September 2025.
Hello, everyone. I observed the more general case where 2 is replaced by 0<|q|<1, a complex algebraic number. Here is the PDF of the comment. The Wikipedia page (in the references) should be consulted in its French version. I think I must have made some kind of mistake, but I can't find it. Feel free to delete my message if you find an error. Thank you.
Note’s PDF
Hi Roman! I don't see where Bertrand proves that $\sum_{n\geq1}q^{n^2}$ is transcendental for algebraic $q$? That paper contains various results ruling out some algebraic relations, but the actual transcendence of that series does not seem to be settled; or am I missing something? EDIT: I was missing something, it is indeed a special case of Theorem 4 in Bertrand that $\sum_{n\geq1}q^{n^2}$ is transcendental for algebraic $0<q<1$, although as he notes, this is actually first proved by Nesterenko.
Even given this, your proof gives that, if $A(q)=\sum_{n\geq 1}\sigma(n)q^{n}$, then $A(q)-4A(q^4)$ is transcendental, for any algebraic $q$. This implies, for example, that at least one of $A(q)$ and $A(q^4)$ is transcendental, but doesn't imply anything about $A(q)$ itself directly.
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