Cauchy–Hadamard theorem

In mathematics, the Cauchy–Hadamard theorem is a result in complex analysis named after the French mathematicians Augustin Louis Cauchy and Jacques Hadamard, describing the radius of convergence of a power series. It was published in 1821 by Cauchy, but remained relatively unknown until Hadamard rediscovered it. Hadamard's first publication of this result was in 1888; he also included it as part of his 1892 Ph.D. thesis.

Theorem for one complex variable
Consider the formal power series in one complex variable z of the form $$f(z) = \sum_{n = 0}^{\infty} c_{n} (z-a)^{n}$$ where $$a, c_n \in \Complex.$$

Then the radius of convergence $$R$$ of f at the point a is given by $$\frac{1}{R} = \limsup_{n \to \infty} \left( | c_{n} |^{1/n} \right)$$ where $lim sup$ denotes the limit superior, the limit as $n$ approaches infinity of the supremum of the sequence values after the nth position. If the sequence values are unbounded so that the $lim sup$ is ∞, then the power series does not converge near $a$, while if the $lim sup$ is 0 then the radius of convergence is ∞, meaning that the series converges on the entire plane.

Proof
Without loss of generality assume that $$a=0$$. We will show first that the power series $\sum_n c_n z^n$ converges for $$|z|R$$.

First suppose $$|z| 0$$, there exists only a finite number of $$n$$ such that $\sqrt[n]{|c_n|} \geq t+\varepsilon$. Now $$|c_n| \leq (t+\varepsilon)^n$$ for all but a finite number of $$c_n$$, so the series $\sum_n c_n z^n$ converges if $$|z| < 1/(t+\varepsilon)$$. This proves the first part.

Conversely, for $$\varepsilon > 0$$, $$|c_n|\geq (t-\varepsilon)^n$$ for infinitely many $$c_n$$, so if $$|z|=1/(t-\varepsilon) > R$$, we see that the series cannot converge because its nth term does not tend to 0.

Theorem for several complex variables
Let $$\alpha$$ be an n-dimensional vector of natural numbers ($$\alpha = (\alpha_1, \cdots, \alpha_n) \in \N^n$$) with $$||\alpha|| = \alpha_1 + \cdots + \alpha_n$$, then $$f(x)$$ converges with radius of convergence $$\rho = (\rho_1, \cdots, \rho_n) \in \R^n$$ with $$\rho^\alpha = \rho_1^{\alpha_1} \cdots \rho_n^{\alpha_n}$$ if and only if $$\limsup_{||\alpha||\to\infty} \sqrt[||\alpha||]{|c_\alpha|\rho^\alpha}=1$$ to the multidimensional power series $$\sum_{\alpha\geq0}c_\alpha(z-a)^\alpha := \sum_{\alpha_1\geq0,\ldots,\alpha_n\geq0}c_{\alpha_1,\ldots,\alpha_n}(z_1-a_1)^{\alpha_1}\cdots(z_n-a_n)^{\alpha_n}$$

Proof
From

Set $$z = a + t\rho$$ $$(z_i = a_i + t\rho_i)$$, then


 * $$\sum_{\alpha \geq 0} c_\alpha (z - a)^\alpha = \sum_{\alpha \geq 0} c_\alpha \rho^\alpha t^{||\alpha||} = \sum_{\mu \geq 0} \left( \sum_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha \right) t^\mu$$

This is a power series in one variable $$t$$ which converges for $$|t| < 1$$ and diverges for $$|t| > 1$$. Therefore, by the Cauchy-Hadamard theorem for one variable


 * $$\limsup_{\mu \to \infty} \sqrt[\mu]{\sum_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha} = 1$$

Setting $$|c_m| \rho^m = \max_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha$$ gives us an estimate


 * $$|c_m| \rho^m \leq \sum_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha \leq (\mu + 1)^n |c_m| \rho^m$$

Because $$\sqrt[\mu]{(\mu + 1)^n} \to 1$$ as $$\mu \to \infty$$


 * $$\sqrt[\mu]{|c_m| \rho^m} \leq \sqrt[\mu]{\sum_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha} \leq \sqrt[\mu]{|c_m| \rho^m} \implies \sqrt[\mu]{\sum_{||\alpha|| = \mu} |c_\alpha| \rho^\alpha} = \sqrt[\mu]{|c_m| \rho^m} \qquad (\mu \to \infty)$$

Therefore


 * $$\limsup_{||\alpha||\to\infty} \sqrt[||\alpha||]{|c_\alpha|\rho^\alpha} = \limsup_{\mu \to \infty} \sqrt[\mu]{|c_m| \rho^m} = 1$$