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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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    Tha$n$ks for your a$n$swers, 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 $m$e 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 $|x|<1$.