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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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    the open unit disk as the function $1/(1-x^2)$. So yes, in this sense they are mathematically equivalent.2011-10-05

3 Answers 3

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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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    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.2011-10-05
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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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    @UGPhysics, right.2011-10-05