# Why is Taylor series expansion for $1/(1-x)$ valid only for $x \in (-1, 1)$?

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". – David Mitra Sep 23 '12 at 18:48
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? – Dávid Tóth Sep 23 '12 at 19:11
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. – Hans Lundmark Sep 23 '12 at 19:39
I see @John got to GH Hardy's classic before I did - and you could look at 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. – Mark Bennet Sep 23 '12 at 19:59

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. - 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\$.