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Let $z_1,...,z_{\ell}$ be complex numbers with $\left|z_k\right|>1$ for all $k=1,...,\ell$. Consider the following sequence

$$ \eta_q = \sum_{\left(i_1,...,i_\ell\right)\in\mathbb N^\ell\atop i_1+...+i_\ell=q}\frac{1}{z_1^{i_1}}\,\frac{1}{z_2^{i_2}}\,\cdots\,\frac{1}{z_\ell^{i_\ell}} $$

Hence, for example, if $\ell=2$

$$ \eta_1 = \frac{1}{z_1}+\frac{1}{z_2}, $$

since $(1,0)$ and $(0,1)$ are the unique couples of integers allowed in the sum. Similarly

$$ \eta_2 = \frac{1}{z_1^2}+\frac{1}{z_2^2}+\frac{1}{z_1\,z_2} . $$

since $(2,0)$, $(0,2)$ and $(1,1)$ are the unique couples of integers allowed in the sum, and so on.

Note that

$$ \eta_q\neq \left(\frac{1}{z_1}+...+\frac{1}{z_{\ell}}\right)^q, $$

in fact, for example (consider again the case $\ell=2$),

$$ \left(\frac{1}{z_1}+\frac{1}{z_{2}}\right)^2 = \frac{1}{z_1^2}+\frac{1}{z_2^2}+\frac{2}{z_1\,z_2}\neq \eta_2. $$

I am pretty sure that the series $\sum_{q=1}^{\infty}\eta_q$ converges thanks to the convergence of all the geometric series

$$ \sum_{q=0}^{\infty}\frac{1}{z_k^q},\quad k=1,...,\ell $$

but I miss a formal argument.

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Isn't your big sum (probably with extra $\eta_0=1$ added) just a product of the geometric series $\sum\limits_{i=0}^\infty\frac{1}{z_k^i}$?

As for the convergence, why, you may get an upper bound for $|\eta_q|$ by changing all $|z_k|$ to the smallest one (which is still $>1$). Then you'll have a fraction with number of those sets $(i_1,\dots,i_l)$ in the numerator and $|z_{min}|^q$ in the denominator. Now, the former is polynomial in $q$ and the latter exponential in it. This proves that the series converges absolutely, so you may pretty well rearrange the terms all you want.

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  • $\begingroup$ so your guess is that $$ 1+\sum_{q=1}^{\infty}\eta_q = \prod_{k}\sum_{i=0}^{\infty}\frac{1}{z_k^i}. $$ This is also my guess, it should be proved with some re-arrangement argument though. Am I right? $\endgroup$ – AlmostSureUser Aug 26 '16 at 14:59
  • $\begingroup$ Yeah, sort of. Well, why even bother with rearrangement. For every $(i_1,...,i_{\ell})\in\mathbb{N}^{\ell}$, your big sum contains $\frac1{z_1^{i_1}}\,\frac{1}{z_2^{i_2}}\,\cdots\,\frac{1}{z_{\ell}^{i_{\ell}}}$ exactly once, and so does my product. $\endgroup$ – Ivan Neretin Aug 26 '16 at 15:09
  • $\begingroup$ thanks, just a last question, how do you say that the cardinality of $$ I=\left\{\left(i_1,...,i_{\ell}\right)\mid i_1+...+i_{\ell}=q\right\} $$ is polynomial in $q$ ? $\endgroup$ – AlmostSureUser Aug 26 '16 at 15:38
  • $\begingroup$ Ok, I think the solution is $$ \frac{(q+\ell-1)!}{q!\,(\ell-1)!} $$ the polynomial growth follows from Stirling's approximation. $\endgroup$ – AlmostSureUser Aug 26 '16 at 16:23
  • $\begingroup$ The polynomial growth follows from ${(q+\ell)!\over q!\;\ell!}={\overbrace{(q+l)\dots(q+1)}^{l\; terms}\over\,l!}$, without any need to resort to anything as complicated as Stirling (which is fairly simple too, and quite useful, but anyway). $\endgroup$ – Ivan Neretin Aug 26 '16 at 19:00

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