# Binomial Expansion to the power of a non-natural number

What is the reasoning that $$|x|$$ has to be less than $$1$$ for $$(1+x)^n$$ when $$n$$ is not a natural number?

• radius of convergence for taylor series. – N. S. Oct 23 '18 at 0:09
• Were |x| > 1, the terms in the expansion would grow without bound. – ncmathsadist Oct 23 '18 at 0:11

## 2 Answers

If you want an intuitive explanation, then consider this scheme.

The Binomial expansion is a Taylor series at $$x=0$$.
The Taylor series is a polynomial that you can view as the polynomial that interpolates "a number" of points close to the origin.

Now, when $$n$$ is an integer, a polynomial of degree $$n$$ interpolating near the origin will also interpolate $$(1+x)^n$$ perfectly for $$x \to \infty$$.

If $$n$$ is not an integer, than for large $$x$$ $$(1+x)^n \approx x^n$$ and no polynomial of integral degree can approximate a non integral power of x.

The radius of convergence $$R$$ of the power series $$\sum_{k \geq 0} \binom{\alpha}{k} x^k$$ is, by the ratio test,

$$R = \lim_{k \to \infty} \left\lvert \frac{\binom{\alpha}{k}}{\binom{\alpha}{k+1}}\right\rvert = \lim_{k \to \infty} \left\lvert\frac{k+1}{\alpha - k}\right\rvert = 1$$

And so we are only guaranteed convergence for those $$x \in \mathbb{C}$$ satisfying $$|x| < 1$$.

Edit: I should add the caveat above that $$\alpha \in \mathbb{C} \setminus \{0, 1, 2, \ldots\}$$. If $$\alpha$$ is a nonnegative integer, then the sequence $$\binom{\alpha}{0}, \binom{\alpha}{1}, \ldots$$ is eventually zero, and the power series is just a polynomial, which converges everywhere.