# Summability of $\sum\limits_q\sum\limits_{ i_1+\cdots+i_\ell=q}\prod\limits_{k=1}^\ell\frac{1}{z_k^{i_k}}$ when $|z_k|>1$ for every $k$

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.

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.
• 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? – AlmostSureUser Aug 26 '16 at 14:59
• 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. – Ivan Neretin Aug 26 '16 at 15:09
• 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 $q$ ? – AlmostSureUser Aug 26 '16 at 15:38
• Ok, I think the solution is $$\frac{(q+\ell-1)!}{q!\,(\ell-1)!}$$ the polynomial growth follows from Stirling's approximation. – AlmostSureUser Aug 26 '16 at 16:23
• 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). – Ivan Neretin Aug 26 '16 at 19:00