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My books states that if the convergence radius of a complex power series is $+\infty$, then the power series is uniformly convergent over every 'disk' of the complex plane, although not necessarily over the entire complex plane. How is this possible?

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  • $\begingroup$ The key is uniformly convergent. $\endgroup$
    – copper.hat
    Jan 5, 2013 at 18:40
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    $\begingroup$ A somewhat oblique analogy: $f(z) = z$ is bounded on every disk, but not over the entire complex plane :) $\endgroup$ Jan 5, 2013 at 19:25

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Take $f(x) = e^x = \sum_{k=0}^\infty \frac{1}{k!} x^k$, $f_N(x) = \sum_{k=0}^N \frac{1}{k!} x^k$. Then the function $x \mapsto f(x)-f_N(x)$ is an entire function, hence unbounded on $\mathbb{C}$. So the convergence is not uniform on $\mathbb{C}$.

Of course, if $|x| \leq R$, then $|f(x)-f_N(x)| \leq \sum_{k=N+1}^\infty \frac{1}{k!} R^k= e^R-\sum_{k=0}^N \frac{1}{k!} R^k$. Since $\lim_{N \to \infty} \sum_{k=0}^N \frac{1}{k!} R^k = e^R$, it follows from the Weierstrass M-test that convergence is uniform on $\overline{B(0,R)}$.

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Consider $$\exp(z) := \sum_{k \geq 0} \frac{z^k}{k!}$$

Let $f_n(z) := \sum_{k=0}^n \frac{z^k}{k!}$ the partial sum of the series. Then

$$\sup_{z \in \mathbb{C}} |f_n(z)-f(z)| \geq \sup_{z \in [0,\infty)} |f_n(z)-f(z)| = \sup_{z \in [0,\infty)} \left| \sum_{k \geq n+1} \frac{z^k}{k!} \right| \geq \sup_{z \in [0,\infty)} \frac{z^{n+1}}{(n+1)!} = \infty$$

i.e. there cannot exist $n \in \mathbb{N}$ such that $\|f_n-f\|_{\infty} \leq \varepsilon$. This means that $f_n$ does not converge uniformly on $\mathbb{C}$ to $\exp$ (although the radius of convergence is $\infty$).

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