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After finding an expansion of $$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots$$ a quick test of various values for $x$ reveals that this expansion is not valid for $\forall x \in \mathbb{R}-\{1\}$.

When $x=2$, then $ \frac{1}{1-x} = -1 $.

But $1 + x + x^2 + x^3 + \ldots = 1 + 2 + 4 + 8 + \ldots > -1$.

What is going on? I checked for divisibility by $0$, but could not find any flaw.

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    Try googling "radius of convergence".2012-09-23
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    http://archive.org/details/divergentseries033523mbp2012-09-23
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    Thanks for the input. What is special about our function that its expansion is not convergent for all reals? Could we deduce certain facts about the properties of the expansion before even attempting to do one?2012-09-23
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    The power series representation is valid in the largest disk (in the complex plane) around the origin which doesn't contain a singular point. In this case, $x=1$ is a singularity (because of division by zero), so the power series is not valid outside the unit disk.2012-09-23
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    I see @John got to GH Hardy's classic before I did - and you could look at http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_… . On the whole, though, the usual answers about radius of convergence are much more useful until you encounter contexts in which the Divergent Series material makes sense.2012-09-23
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    @DavidToth Physicists have built whole worlds around this. Some of them are really similar to our.2012-09-23
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    Thanks for your answers, I actually knew about the radius of convergence and that the given series was divergent - I should have asked in a better way that I did not understand why the given series had that property while other functions when expanded have convergent series for all the values. Hardy's book gives me a good direction further.2012-09-23

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The following is valid for all $x$$$1+x+x^2+\cdots +x^n=\frac{1-x^{n+1}}{1-x}$$ Now taking limit as $n\to\infty $ you see that $x^{n+1}\to 0$ only when $|x|<1$ and that gives your formula.

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Well, we have that

$$s_n=\sum_{k=0}^{n}x^k=\frac{1-x^{n+1}}{1-x}$$

Now, we look at $n\to \infty$

$$\lim_{n\to\infty}\frac{1-x^{n+1}}{1-x}=\lim_{n\to\infty}\sum_{k=0}^{n}x^k$$

For what $x$ does this limit exist?

Clearly, we're only concerned about $x^{n+1}$. If $|x|>1$, then $x^{n+1}$ grows large, and there is no limit. If $|x|<1$, then $x-1\neq 0$ and $x^{n+1}$ goes to $0$ as $n\to\infty$, so in that case, we can assert that

$$\lim_{n\to\infty}\frac{1-x^{n+1}}{1-x}=\frac 1{1-x}=\sum_{k=0}^{\infty}x^k$$

If $x=1$, $x-1=0$ and we find ourselves in trouble. However, we can say that $$\sum\limits_{k = 0}^n {{1^k}} = n$$ in which case the sequence of partial sums has no limits. Finally, for $x=-1$, we have $(-1)^{n+1}$ which oscillates and has no limit.

Thus, $$\frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty {{x^k}} $$ makes sense only for $|x|<1$.