Unlike polynomials power series do not grow uniformly. For instance, if $n \geq 1$,
\[
P(z) = a_n z^n + \cdots + a_1 z + a_0,
\]
and $a_n \neq 0$, then $P(z) \sim a_n z^n$. In particular, $|P(z)| \to \infty$ as $|z| \to \infty$. But a power series, such as sine or cosine, can have zeros tending to $\infty$. However, power series do grow on average. We explore this below.
Let $f(z) = \sum_{n = 0}^{\infty} a_n (z-a)^n$, $z \in D(a, R)$. Then for $0 \leq r < R$ we have
\begin{align*}
\int_{0}^{2 \pi} |f(a + r e^{i \theta})|^2 \, d \theta &= \int_{0}^{2 \pi} \sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty} a_n \overline{a_m} r^{n + m} e^{i(n- m) \theta} \, d \theta \\ &= \sum_{n = 0}^{\infty} \sum_{m = 0}^{\infty} a_n \overline{a_m} r^{n + m} \int_{0}^{2 \pi } e^{i (n-m) \theta} \, d \theta \\ &= 2 \pi \sum_{n = 0}^{\infty} |a_n|^2 r^{2n}
\end{align*} Note that we can switch the order of summation and integration since the double sum converges absolutely and uniformly (due to Weierstrass $M$-test) as a function of $\theta$. Thus we have
\[
\frac{1}{2 \pi} \int_{0}^{2 \pi} |f(a + r e^{i \theta})|^2 \, d \theta = \sum_{n = 0}^{\infty} |a_n|^2 r^{2n}.
\] This is called Parseval’s identity. We now employ this formula to give a proof of an important result in complex analysis.
Theorem. (Liouville’s theorem) Let $f$ be a bounded and entire. Then $f$ must be constant.
Proof. Let $f$ be a bounded and entire function with $|f(z)| \leq M$ for every $z \in \mathbb{C}$. Since $f$ is entire we can express it as a power series throughout $\mathbb{C}$ as
\[
f(z) = \sum_{n = 0}^{\infty} a_n z^n.
\] It now follows by Parseval’s identity that
\[
\sum_{n = 0}^{\infty} |a_n|^2 r^{2n} = \frac{1}{2 \pi} \int_{0}^{2 \pi } |f(re^{i \theta})|^2 \, d \theta \leq M^2
\] for every $r > 0$. Therefore we have
\[
|a_n|^2 \leq \frac{M^2}{r^{2n}}
\] for each $n \geq 1$. It now follows by letting $r$ tend to $\infty$ that $a_n = 0$ for $n \geq 1$ and hence $f$ must be constant.