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By trying to prove that Riemann's Zeta function is analytically expendable to the whole plane with one pole, I went aside and noticed this identity about formal power series (which are obviously everywhere convergent for complex $x$): $$ \prod_{n=0}^{\infty}\left(1+\frac{x}{a^n}\right)=\sum_{n=0}^{\infty}\frac{(ax)^n}{\prod_{k=1}^{n}(a^k-1)},\quad\quad\forall a\in\mathbb{R}:a>1. $$ I checked it for the first several terms on my computer and it matches. Can you help me prove this identity? In my proof I would plug in $a=4$ and $a=3$ to prove that the Zeta function has the domain it has, although my first example to check was $a=2$. My idea was to continue Ramanujan's idea of transforming Riemann's Zeta function into Dirichlet's Eta function to represent Zeta as a quotient of two Dirichlet series that converge everywhere, denominator would be the function in the identity, with $x=2^{-s}$. That's the idea, the proof is not finished, so I am aware that this idea will probably fail in this shape. I will have to work more on it.

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    $\begingroup$ When you say you “noticed” this identity, and your question asks us to prove the identity; how do you even know for sure it is an identity? Where did it come from? The work of another author? $\endgroup$
    – FShrike
    Dec 21, 2021 at 23:19
  • $\begingroup$ @FShrike I checked first few coefficients on computer $\endgroup$
    – donaastor
    Dec 21, 2021 at 23:21
  • $\begingroup$ For all real $p$, $p\gt1$, and yet you cite specific values of $p$ less than one. What also is the domain of $x$? $\endgroup$
    – FShrike
    Dec 21, 2021 at 23:22
  • $\begingroup$ Yes, sorry. I written the identity wrongly the first time, but when I try to correct it it is a race against other edits. Sorry, come back to the link in few minutes $\endgroup$
    – donaastor
    Dec 21, 2021 at 23:25
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    $\begingroup$ @Jacob Thank you very much! Would you post it as an answer so I can mark it and close the question? $\endgroup$
    – donaastor
    Dec 22, 2021 at 1:59

2 Answers 2

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For $|a|>1$ let $$f(x)=\prod_{n=0}^\infty (1+x a^{-n}) =\sum_{m=0}^\infty c_m x^m$$ $$\sum_{m=0}^\infty c_m a^m x^m=f(ax)= (1+x a) f(x)=1+\sum_{m=1}^\infty (c_m+a c_{m-1})x^m$$ Equating the coefficients we find that $$c_0=1,\qquad c_m a^m=c_m +a c_{m-1}$$ ie. $$c_m = \frac{a}{a^m-1}c_{m-1} = \frac{a^m}{\prod_{k=1}^m (a^k-1)}$$

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In accordance with OP’s request, I am posting my comment below.

The identity is a theorem of Euler’s that comes from the $q$-binomial theorem in the limit $n\to\infty$. A proof can be found here.

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  • $\begingroup$ See my answer, no need of any theorem $\endgroup$
    – reuns
    Dec 22, 2021 at 5:13

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