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Given the geometric series:

$1 + x^2 + x^4 + x^6 + x^8 + \cdots$

We can recast it as:

$S = 1 + x^2 \, (1 + x^2 + x^4 + x^6 + x^8 + \cdots)$, where $S = 1 + x^2 + x^4 + x^6 + x^8 + \cdots$.

This recasting is possible only because there is an infinite number of terms in $S$.

Exactly how is this mathematically possible?

(Related, but not identical, question: General question on relation between infinite series and complex numbers).

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Are you asking why $\frac{1}{1-x^2} = 1 + x^2 \frac{1}{1-x^2}$? Note that the identification of the formal series $S$ with the function $\frac{1}{1-x^2}$ is valid for $|x| \lt 1$, otherwise it's just a game you can play with formal power series. –  t.b. Oct 5 '11 at 12:31
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+1 For realizing that it is essential for there to be an infinite number of terms. Infinite sums usually make sense only under some kind of convergence (in a ring of formal power series the convergence is of a very different kind). But yeah, it is the same phenomenon that is underlying the following. Let $S=0.9999\ldots$. Then $$S=0.9+0.09999\ldots=0.9+0.1\cdot S,$$ Therefore $0.9=S-0.1\cdot S=0.9 \cdot S$, so $S=0.9/0.9=1$. –  Jyrki Lahtonen Oct 5 '11 at 12:40
    
Well, I'm basically trying to understand why we can do this recasting. Normally - (I'm assuming) - we cannot recast a finite series. But the above recasting is allowed since the terms go to infinity. ... What is the exact mathematical (i.e. rigorous) basis which allows this manipulation? –  UGPhysics Oct 5 '11 at 12:47
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You cannot even define the sum of infinitely many elements without some notion of a convergence. In the ring of formal power series (that many of us want to use here!) the convergence is based on the so called $x$-adic topology. Meaning that an infinite sum $\sum_{i=0}^\infty p_i(x)$ makes sense, if and only if the degrees of the lowest degree terms of the power series $p_i(x)$ tend to infinity. Here $p_i(x)=x^{2i}$ is its lowest degree term, and convergence thus follows from the fact that $2i\to\infty$ as $i\to\infty$. –  Jyrki Lahtonen Oct 5 '11 at 13:18
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IOW those three dots $\ldots$ hide a lot of reasoning. And 'no', the recasting does not make sense without some kind of a convergence allowing infinite sums to be formed in whatever structure we feel is most appropriate (here the complex numbers or the ring of formal power series). –  Jyrki Lahtonen Oct 5 '11 at 13:22
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3 Answers

There is a finite version of which the expression you have is the limit.

Suppose $S=1+x^2+x^4+x^6+x^8$, then we can put

$S+x^{10}=1+x^2(1+x^2+x^4+x^6+x^8)=1+x^2S$

And obviously this can be taken as far as you like, so you can replace 10 with 10,000 if you choose. If the absolute value of $x$ is less than 1, this extra term approaches zero as the exponent increases.

There is also a theory of formal power series, which does not depend on notions of convergence.

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If x is a matrix, nilpotent to some even power, can we then say, that this holds also for finite series? (Or, more generally, if we have an algebra with zero-divisors, can we formally relax the condition of infiniteness?) –  Gottfried Helms Oct 5 '11 at 12:56
    
Hi @Mark, thanks for the response. | May I know the formal power series version of your response? (I think that's what I'm searching for.) –  UGPhysics Oct 5 '11 at 13:21
    
lhf has put a link to some notes on formal power series, which covers more than any comment I could make. What I wanted to show was that, while it is not possible to recast a finite truncation of your series in the form you gave, it is possible to get close if you include a kind of error term. Roughly, in formal power series the powers of $x$ act as placeholders for the coefficients. Arithmetic operations work as you would expect, but you have to be careful that the calculation for each coefficient of the result is a finite calculation. –  Mark Bennet Oct 5 '11 at 13:55
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The $n$th partial sum of your series is

$$ \begin{align*} S_n &= 1+x^2+x^4+\cdots +x^{2n}= 1+x^2(1+x^2+x^4+\cdots +x^{2n-2})\\ &= 1+x^2S_{n-1} \end{align*} $$

Assuming your series converges you get that $$ \lim_{n\to\infty}S_n=\lim_{n\to\infty}S_{n-1}=S. $$

Thus $S=1+x^2S$.

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$S = 1 + x^2 \, S$ is true even in the ring of formal power series. No convergence is needed here.

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If I understand correctly, the series in the question can be recast only if it is convergent; otherwise it can't. [via @Jyrki | ... so, how then are "formal power series" related to the "usual" series above? ... –  UGPhysics Oct 5 '11 at 16:36
    
@UGPhysics, if you want to talk about numerical series, where $x$ is a number, then of course you need to worry about convergence. The manipulation you mentioned in the question is really purely formal. That was my point. –  lhf Oct 5 '11 at 17:58
    
So, -to clarify this whole debate-, in the space of real numbers I cannot write an expression like $S$, and claim that I can recast it [as $1 + x^2\,S$] unless the expression is convergent, (or defined within a specified radius of convergence)? –  UGPhysics Oct 5 '11 at 18:06
    
@UGPhysics, right. –  lhf Oct 5 '11 at 19:44
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