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For $x > 1$, the geometric series $\sum\limits_{i = 0}^n{x}^i$ is equal to $\frac{x^{n+1}-1}{x-1}$.

By getting the limit,

$$ \lim\limits_{n \rightarrow \infty} \sum\limits_{i = 0}^n{x}^i = \lim\limits_{n \rightarrow \infty} \frac{x^{n+1}-1}{x-1} $$

$$ = \infty $$

However, the generating function $\sum\limits_{n \geq 0}x^n$ is equal to $\frac{1}{1-x}$ since

$$ \sum\limits_{n \geq 0}{x}^n - x\sum\limits_{n \geq 0}{x}^n= (1 + x + x^2 + x^3 + \dots) - (x + x^2 + x^3 + \dots) = 1 $$

$$ \Rightarrow (1-x)\sum\limits_{n \geq 0}{x}^n= 1 $$

$$ \Rightarrow \sum\limits_{n \geq 0}{x}^n= \frac{1}{1-x} $$

Obviously, $\frac{1}{1-x}$ is a finite number.

What is the underlying concept that makes them yield different results?

edgar
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3 Answers3

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Although this is not the intended interpretation, one could also make sense of the identity $$ \sum_{n \geq 0} x^n = (1-x)^{-1} \tag{$\dagger$} $$ formally; i.e., in the ring of formal power series $\mathbf C[[x]]$. The user Bill Dubuque explains the idea in several posts in this site; see, e.g., his answer to the post Product of two power series.

The advantage in interpreting $(\dagger)$ formally is that since $x$ is an indeterminate rather than a number, there are no longer any convergence issues. In this sense, the series $\sum_{n \geq 0} x^n$ is the same as the finite thing $(1-x)^{-1}$.

The flip side of this is that one can no longer plug in values into the series: it does not make sense to talk about the sum of the series at $x = \frac{1}{2}$. Thus the statement "the series converges for $|x| < 1$ and diverges otherwise" is meaningless under this interpretation.

Community
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Srivatsan
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    (I am fully aware that there is a chance the OP may not understand the answer. This post is meant for all readers, not just the OP.) – Srivatsan Feb 01 '12 at 16:47
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They're not different. The generating function is only defined for when $|x| < 1$. Similarly, the geometric series only converges for $|x| < 1$. And in that case, $\lim x^n = 0$ and you see that the two expressions are the same.

davidlowryduda
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The given derivation of the fact that $\sum_{ n \geq 0 } x^n = \frac{ 1 }{ 1 - x }$, doesn't address convergence of the intermediate series at all. You can't ignore this issue in a derivation and then hope to plug in numbers to the result.

The derivation given shows that the given identity is valid in the context of formal power series, which are important in the study of generating functions.

As @mixedmath correctly states, you may only substitute numbers into this identity if $| x | < 1$.

Austin Rochford
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