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I'm studying physics and am currently following a course on complex analysis and in the section on analytic functions, the radius of convergence $R$ for power series was introduced. The Taylor expansion around $z_0=0$ for the exponential function was considered as an example of a power series with $R\rightarrow\infty$. The notes state this can be proved by using Weierstrass' Criterion for uniform convergence, which I'll state in my own words:

Consider a series

$\sum\limits_{k=0}^{\infty} f_k(z)$.

If you know numbers $a_k$ for which

$|f_k(z)| < a_k$

for all $z$, and

$\sum\limits_{k=0}^{\infty} a_k$

converges uniformly, then also

$\sum\limits_{k=0}^{\infty} f_k(z)$

converges uniformly.

For the exponential, we have the power series

$e^z = \sum\limits_{k=0}^{\infty}\dfrac{z^k}{k!}$.

Now I've been thinking about this, but I can't seem to think of a uniformly converging series of $a_k$'s that bound the terms of this power series. Perhaps this is really straightforward and I wouldn't have any difficulties with it if I remembered my course on real analysis a bit better...
It's not a homework problem and series convergence is not a main goal in this course, but it's been bugging me that I don't understand why Weierstrass's Criterion proves that the radius of convergence goes to infinity for the exponential, so I thought I'd ask here. Thanks in advance.

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  • $\begingroup$ The constants $a_k$ depend on $R$. You need to show that you can find the sequence $a_k$ for each given $R$. (I'm guessing you're trying to find one sequence of $a_k$ that works for all $R$. That's not possible.) $\endgroup$
    – Tunococ
    Jan 19, 2013 at 13:16

1 Answer 1

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You need to be careful in making the distinction between uniform convergence for any $z \in \mathbb{C}$ and uniform convergence for $|z| < R$ for a fixed $R$.

The exponential function is unbounded. If we were able to find a sequence $a_k$ so that for any $z \in \mathbb{C}$ $$ \left|e^z\right| = \left|\sum_{k=0}^\infty \dfrac{z^k}{k!}\right| \le \sum_{k=0}^\infty \left|\dfrac{z^k}{k!}\right| \le \sum_{k=0}^\infty a_k < \infty $$

then the exponential function would be bounded, a contradiction.

Now, if we limit the domain of the exponential function to $|z| < R$ for a fixed $R > 0$, then: $$ \sum_{k=0}^\infty \left|\dfrac{z^k}{k!}\right| \le \sum_{k=0}^\infty \dfrac{R^k}{k!} < \infty $$

The convergence of $\displaystyle \sum_{k=0}^\infty \frac{R^k}{k!}$ can be established via the ratio test. Thus, the exponential function is uniformly convergent for any fixed $R > 0$ no matter how big it is.

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  • $\begingroup$ Ahh, beautiful :) Thank you for this clear explanation, @Tunococ was right that I was looking for one sequence of $a_k$ for all $R$. $\endgroup$
    – Wouter
    Jan 19, 2013 at 13:35

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