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.
$\endgroup$ 72 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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 $|x|<1$.
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