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

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

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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.

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