# Analytic Function without Power Series

If $$f(x) =\sum_{n=0}^{\infty}x^ n$$ Then Determine the function $f(x)$. Discuss the domain of $f(x)$. Discuss the domain of the derivative of $f(x)$. Thanks!

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What is $x_n$? Do you mean $x^n$? – Robert Israel Feb 7 '13 at 7:23
@RobertIsrael: The OP wrote x n. – Babak S. Feb 7 '13 at 7:35
Do you know about geometric series? – Daenerys Naharis Feb 7 '13 at 7:48

If $x = 1$, then the infinite series diverges to $\infty$. Hence, suppose that $x \in \mathbb{R} \setminus \{ 1 \}$. Then for all $N \in \mathbb{N}$, we have $$\sum_{n=0}^{N} x^{n} = \frac{1 - x^{N+1}}{1 - x}.$$ The sum on the left-hand side of this identity converges as $N \to \infty$ if and only if the sequence $(x^{N+1})_{N \in \mathbb{N}}$ converges if and only if $|x| < 1$. Therefore, $\text{Dom}(f) = (-1,1)$, and $$\forall x \in (-1,1): \quad f(x) = \lim_{N \to \infty} \frac{1 - x^{N+1}}{1 - x} = \frac{1}{1 - x}.$$ Finally, observe that by either the Quotient Rule or the Chain Rule, $f$ is differentiable on $(-1,1)$.
I’m not sure what you mean, so the best that I can say is that $f(x)$ is undefined if $x \leq -1$, is $\dfrac{1}{1 - x}$ if $-1 < x < 1$ and is $+ \infty$ if $1 \leq x$. – Haskell Curry Feb 7 '13 at 9:52
It is necessary to first consider the finite sum $\displaystyle \sum_{n=0}^{N} x^{n}$ because $\displaystyle \sum_{n=0}^{\infty} x^{n} \stackrel{\text{def}}{=} \lim_{N \to \infty} \sum_{n=0}^{N} x^{n}$. – Haskell Curry Feb 7 '13 at 9:56