Cauchy-Hadamard Theorem/Complex Case

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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:

$\ds \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$


$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$

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:

$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$


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:

$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$



$\ds \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$

Hence the result.


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 $\tilde r = r \paren {L + \epsilon}$.


$0 \le \tilde r < 1$

This means that the geometric series $\ds \sum_{n \mathop = 0}^\infty$ is convergent.

From the properties of the limit superior, we know that:

$\forall n \gg 1: \cmod {a_n}^{1/n} < L + \epsilon$


$\forall n \gg 1: \cmod {a_n r^n} < \paren {L + \epsilon}^n r^n = \tilde r^n$

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 $r \in \closedint 0 {\dfrac 1 L}$.


$\dfrac 1 L \le R$

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 $k \in \N$.


$\cmod {a_{n_k} \cdot \dfrac 1 {\paren {L + \epsilon}^{n_k} } } > 1$

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.


$R \le \dfrac 1 {L + \epsilon}$

This holds for all $\epsilon \in \R_{>0}$.


$R \le \dfrac 1 L$


$\dfrac 1 L \le R \le \dfrac 1 L$

which means:

$R = \dfrac 1 L$

as required.


Source of Name

This entry was named for Augustin Louis Cauchy and Jacques Salomon Hadamard.


  • 1888: Jacques HadamardSur le rayon de convergence des séries ordonnées suivant les puissances d'une variable (C.R. Acad. Sci. Vol. 106: pp. 259 – 262)