关于解析函数收敛和及导数级数收敛推整函数的证明问询

Hey there! Let's break this down step by step since you already noticed the key starting point that $|f^{(n)}(z_0)| \to 0$ as $n \to \infty$.

First, remember that for an analytic function $f(z)$ near $z_0$, its Taylor series centered at $z_0$ is:
$$f(z) = \sum_{n=0}^{\infty} \frac{f^{{(n)}(z_0)}{n!}(z-z_0)}n$$
The radius of convergence $R$ of this series is given by the Cauchy-Hadamard formula:
$$\frac{1}{R} = \limsup_{n \to \infty} \left| \frac{f^{(n)}(z_0)}{n!} \right|^{1/n}$$

Now, since the series $\sum_{n=1}^{\infty} f^{(n)}(z_0)$ converges, the general term must tend to 0 (as you noted), but more importantly, there exists some constant $M > 0$ such that $|f^{(n)}(z_0)| \leq M$ for all sufficiently large $n$—this is a basic property of convergent sequences: they're always bounded.

Let's analyze the term inside the limsup:
$$\left| \frac{f^{(n)}(z_0)}{n!} \right|^{1/n} \leq \frac{M^{1/n}}{(n!){1/n}}$$
We can use two key facts here:

• $M^{1/n} \to 1$ as $n \to \infty$ (any positive constant raised to the power of $1/n$ approaches 1)
• Using Stirling's approximation, we know $n! \sim \sqrt{2\pi n} \left( \frac{n}{e} \right)^n$ for large $n$. Taking the $n$-th root of both sides gives $(n!)^{1/n} \sim \frac{n}{e}$, which tends to $\infty$ as $n \to \infty$.

Putting these together, the right-hand side of the inequality tends to 0 as $n \to \infty$. This means:
$$\limsup_{n \to \infty} \left| \frac{f^{(n)}(z_0)}{n!} \right|^{1/n} = 0$$
Therefore, $\frac{1}{R} = 0$, so $R = \infty$.

This tells us the Taylor series of $f(z)$ centered at $z_0$ converges everywhere on the complex plane. Since $f(z)$ is equal to its Taylor series in its original neighborhood of analyticity, and now that convergence extends to the entire plane, $f(z)$ must be an entire function.

That's the core of the argument—using the boundedness of the derivatives (from the convergent series) and the Cauchy-Hadamard formula to show the Taylor series has infinite radius of convergence.

备注：内容来源于stack exchange，提问作者Relure