Cauchy-Hadamard Theorem/Complex Case
Theorem
Let $\xi \in \C$ be a complex number.
Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.
Then the radius of convergence $R$ of $\map S z$ is given by:
- $\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$
If:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$
then the radius of convergence is infinite, and $\map S z$ is absolutely convergent for all $z \in \C$.
Proof 1
Let $\epsilon \in \R_{>0}$, and let $z \in \C$.
Suppose that $\cmod {z - \xi} = R - \epsilon$.
By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent.
From the $n$th Root Test:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \le 1$
By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$
$\Box$
Now, suppose that $\cmod {z - \xi} = R + \epsilon$.
Then $S \paren z$ is divergent, so the $n$th Root Test shows that:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \ge 1$
which we can rearrange to obtain:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$
As $\epsilon > 0$ was arbitrary, it follows that:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$
$\Box$
Thus:
- $\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$
Hence the result.
$\blacksquare$
Proof 2
Let $L = \limsup \cmod {a_n}^{1/n}$.
We will consider only the case $0 < L < \infty$, as the cases $L = 0$ and $L = \infty$ follow quite simply from this one.
We have that:
- $\forall r \in \closedint 0 {\dfrac 1 L}: L < \dfrac 1 r$
Thus there exists $\epsilon \in \R_{> 0}$ such that:
- $L + \epsilon < \dfrac 1 R$
and so:
- $r < \dfrac 1 {L + \epsilon}$
Let $\tilde r = r \paren {L + \epsilon}$.
Then:
- $0 \le \tilde r < 1$
This means that the geometric series $\ds \sum_{n \mathop = 0}^\infty$ is convergent.
From the properties of the limit superior, we know that:
- $\forall n \gg 1: \cmod {a_n}^{1/n} < L + \epsilon$
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Thus:
- $\forall n \gg 1: \cmod {a_n r^n} < \paren {L + \epsilon}^n r^n = \tilde r^n$
This means that the series $\ds \sum_{n \mathop = 0}^\infty \cmod {a_n r^n}$ is also convergent.
Thus $\ds \sum_{n \mathop = 0}^\infty a_n r^n$ is absolutely convergent and therefore convergent.
Thus, by the definition of radius of convergence, we have:
- $r \le R$
which holds for all $r \in \closedint 0 {\dfrac 1 L}$.
Hence:
- $\dfrac 1 L \le R$
Let now $\epsilon \in \openint 0 L$ be fixed.
Because of the properties of the limit superior, we can extract from the sequence of the formula a subsequence such that:
- $\cmod {a_{n_k} }^{1 / n_k} > L - \epsilon$
for all $k \in \N$.
Thus:
- $\cmod {a_{n_k} \cdot \dfrac 1 {\paren {L + \epsilon}^{n_k} } } > 1$
for all $k \in \N$.
So the series:
- $\ds \sum_{n \mathop = 0}^\infty a_n \paren {\frac 1 {L + \epsilon} }^n$
is not convergent.
Thus:
- $R \le \dfrac 1 {L + \epsilon}$
This holds for all $\epsilon \in \R_{>0}$.
So:
- $R \le \dfrac 1 L$
Hence:
- $\dfrac 1 L \le R \le \dfrac 1 L$
which means:
- $R = \dfrac 1 L$
as required.
$\blacksquare$
Source of Name
This entry was named for Augustin Louis Cauchy and Jacques Salomon Hadamard.
Sources
- 1888: Jacques Hadamard: Sur le rayon de convergence des séries ordonnées suivant les puissances d'une variable (C.R. Acad. Sci. Vol. 106: pp. 259 – 262)