2.5 Abel's Limit Theorem

Abel’s Limit Theorem (Stolz Angle)

Let f (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}

with radius of convergence $R = 1$
Theorem.

If $\sum_{n = 0}^{\infty} a_{n}$ converges, then

lim_{z \to 1} f (z) = f (1) = \sum_{n = 0}^{\infty} a_{n}

whenever $\frac{| 1 - z |}{1 - | z |}$ remains bounded as

z \to 1

(i.e. $z$ approaches $1$ inside a Stolz angle).

We may assume

\sum_{n = 0}^{\infty} a_{n} = 0

since adding a constant to $a_{0}$ doesn’t affect the limit. Define

s_{n} = a_{0} + a_{1} + \dots + a_{n}

\begin{aligned} s_{n} (z) & = a_{0} + a_{1} z + a_{2} z^{2} + \dots + a_{n} z^{n} \\ = \sum_{k = 0}^{n} a_{k} z^{k} = \sum_{k = 0}^{n} (s_{k} - s_{k - 1}) z^{k} (s_{- 1} = 0) \\ = \sum_{k = 0}^{n} s_{k} z^{k} - \sum_{k = 0}^{n} s_{k - 1} z^{k} = \sum_{k = 0}^{n} s_{k} z^{k} - z \sum_{j = 0}^{n - 1} s_{j} z^{j} \\ = s_{n} z^{n} + (1 - z) \sum_{j = 0}^{n - 1} s_{j} z^{j} \\ = (1 - z) (s_{0} + s_{1} z + \dots + s_{n - 1} z^{n - 1}) + s_{n} z^{n} \end{aligned}

Since $s_{n} \to 0$ and $| z | \leq 1$ we have

s_{n} z^{n} \to 0

Therefore,

f (z) = (1 - z) \sum_{n = 0}^{\infty} s_{n} z^{n}

Assume

| 1 - z | \leq K (1 - | z |)

Pick $m$ such that $| s_{n} | < ε$ for all $n \geq m$

\begin{aligned} | \sum_{n = m}^{\infty} s_{n} z^{n} | & \leq ε \frac{| z |^{m}}{1 - | z |} \\ | 1 - z | | \sum_{n = m}^{\infty} s_{n} z^{n} | & \leq | 1 - z |, ε \frac{| z |^{m}}{1 - | z |} \\ \leq K (1 - | z |), ε \frac{| z |^{m}}{1 - | z |} since | 1 - z | \leq K (1 - | z |) \\ = K ε, | z |^{m} \\ \leq K ε \end{aligned}

\begin{aligned} f (z) & = (1 - z) \sum_{n = 0}^{\infty} s_{n} z^{n} \\ = (1 - z) (\sum_{n = 0}^{m - 1} s_{n} z^{n} + \sum_{n = m}^{\infty} s_{n} z^{n}) \\ | f (z) | & \leq | 1 - z | \sum_{n = 0}^{m - 1} | s_{n} | | z |^{n} + | 1 - z | | \sum_{n = m}^{\infty} s_{n} z^{n} | \\ \leq | 1 - z | \sum_{n = 0}^{m - 1} | s_{n} | | z |^{n} + K ε \end{aligned}

The finite sum vanishes as $z \to 1$ because of the factor $| 1 - z |$ and the tail is bounded by $K ε$

Since $ε$ is arbitrary,

lim_{z \to 1} f (z) = 0

Undoing the normalization,

lim_{z \to 1} f (z) = f (1)

Summary of logic:

\sum a_{n} = 0