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In the ring of formal power series $\Bbb R[[x]]$ it is easy to check by induction that $$ 1 = (1-x)(1 + x + x^2 + \cdots). $$ Does this derivation imply the same identity for those real or complex numbers for which all these power series converge?

Basically I want to know how one can show that the intuitive evaluation map is multiplicative on the subset (subring?) of convergent power series.

My formalism: Fix some $z\in\Bbb C$, let $\phi_z : \Bbb R[x] \to \Bbb C$ be the evaluation homomorphism $p(x)\mapsto p(z)$. For a power series $f$ denote by $f^n$ the polynomial obtained from $f$ by truncating after $n$ terms. Then I'd want to extend the definition of $\phi_z$ to power series by setting $\phi_z(f) := \lim_{n\to\infty} \phi_z(f^n)$, which is the limit of partial sums, if this exists.

I would like to see why $\phi_z(fg) = \phi_z(f)\phi_z(g)$ when all of $f$, $g$, and $fg$ converge at $z$.

[My problem is that $(fg)^n$ is not the same as $f^ng^n$. Addition is nicer since $(f+g)^n = f^n + g^n$.]

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    $\begingroup$ As long as you only require conditional convergence rather than absolute convergence, these convergent power series don't form a subring. See en.wikipedia.org/wiki/… . In the case of absolute convergence, everything is easy using the fact that the terms of an absolutely convergent countable sum can be arbitrarily reordered without changing the value. $\endgroup$ – darij grinberg Aug 19 '13 at 10:38
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    $\begingroup$ Fun trivia: $\phi_a(f\circ g)=\phi_{\phi_a(g)}(f)$ doesn't hold in the $p$-adics. $\endgroup$ – anon Aug 19 '13 at 10:40
  • $\begingroup$ Can it happen that $f$ and $g$ converge conditionally, the product $fg$ also converges, but still $\phi_z(fg) \neq \phi_z(f)\phi_z(g)$? The wikipedia example only shows that $fg$ might diverge. $\endgroup$ – Dominik Peters Aug 19 '13 at 11:31

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