Cauchy-Hadamard Theorem/Complex Case/Proof 2

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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

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:

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


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.