Suppose we define the infinite product $\displaystyle \prod_{n=1}^{\infty} (1+a^{-ns})^{a_n}$, where $a_n$ is some given sequence of positive integers. Is there a way, supposing there is a pole at $s=1$, to compute the residue using the $a_n$?
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$\begingroup$ Do we need to choose a value for $a$? $\endgroup$– S. Carnahan ♦Commented Oct 7, 2013 at 22:52
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$\begingroup$ $a$ is some fixed integer. $\endgroup$– Vlad MateiCommented Oct 8, 2013 at 0:24
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$\begingroup$ Well, I know that what I am going to say is trivial, but be it. But if there's a pole at $s=1$, then there exists $n_0 \in \mathbb{N}$ such that: $a_{n_0} <0$, and $(1+a^{-n_0})=0$. From this you find what's this $n_0$ is exactly (upto modulo $2\pi$. $a= e^{i(-\pi/n_0 + 2\pi k/n_0}$. I am a bit rusty with complex calculus, but this might help. Obviously $a\neq 0$. $\endgroup$– AlanCommented Oct 8, 2013 at 0:50
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$\begingroup$ Actually I should have stated that my sequence is formed by positive integers $\endgroup$– Vlad MateiCommented Oct 8, 2013 at 1:01
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$\begingroup$ Presumably what you mean is that the infinite product converges for $s$ in some domain, and this has an analytic continuation to a meromorphic function on a larger domain which has a pole at $s=1$. $\endgroup$– Robert IsraelCommented Oct 8, 2013 at 1:43
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1 Answer
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An example is easy. Denote $z=a^{-(s-1)}$ and take the logarithm. Then $$ \log f(z)=\sum_n a_n a^{-n}z^n+\sum_n a_n [\log(1+a^{-n}z^n)-a^{-n}z^n]\,. $$ Assume that $a_na^{-n}=\frac 1n+O(b^{-n})$ with some $b>1$. Then, since the difference in the brackets is at most $a^{-2n}|z|^{2n}$ in absolute value for $z$ close to $1$, we see that $\log f(z)$ differs from $-\log(1-z)$ by a series converging uniformly in a larger disk.
However, this also shows that the problem is pretty much as hard as determining the value of $g(z)=\sum_n c_nz^n$ at $1$ for a function that has an analytic extension from the unit disk to a neighborhood of $1$. We can write it as a formal limit and even use some summation methods to ensure controlled speed of convergence, of course, but there seems to be little value in the discussion of the issue in such generality. Once we know more details about $a$ and $a_n$, some more useful answer may be possible.
