## Abel’s Theorems for Infinite Series and Applications

Theorem 1. (Abel’s test)

Let $(a_{n})_{n=0}^{\infty}$ be a decreasing sequence of positive real numbers such that $\lim_{n \rightarrow \infty} a_{n} = 0$ and the series

$\displaystyle f(z) := \sum_{j=0}^{\infty}a_{j}z^{j}$

converges for $\left|z\right| < 1$ and diverges for $\left|z\right| > 1$. Then the power series converges for all $z \in \partial\mathbb{D} \setminus \left\{1\right\}$.

Proof. Let $z = e^{i\theta} \in \partial\mathbb{D}$ with $z \neq 1$. For the functions $w^{\frac{1}{2}}$ and $w^{-\frac{1}{2}}$, take the principal branch. Then

$\displaystyle z^{\frac{1}{2}} - z^{-\frac{1}{2}} = e^{i\frac{\theta}{2}} - e^{-i\frac{\theta}{2}} = 2i\sin\left(\frac{\theta}{2}\right) \neq 0$

Let $N > M \in \mathbb{Z}^{\geq 0}$, and denote the $N^{th}$ and $M^{th}$ partial sums by $S_{N} = \sum_{j=0}^{N}a_{j}z^{j}$ and $S_{M} = \sum_{j=0}^{M}a_{j}z^{j}$, respectively. Then

$\begin{array}{lcl}2i\sin\left(\frac{\theta}{2}\right)\left[S_{N} - S_{M}\right] &=& 2i\sin\left(\frac{\theta}{2}\right)\sum_{j=M+1}^{N}a_{j}z^{j}\\ &=& \sum_{j=M+1}^{N}a_{j}\left(z^{j+\frac{1}{2}} - z^{j-\frac{1}{2}}\right)\\&=&\sum_{j=M+2}^{N}\left(a_{j-1} - a_{j} \right)z^{j-\frac{1}{2}} - a_{M+1}z^{M+\frac{1}{2}} + a_{N}z^{N+\frac{1}{2}}\end{array}$

Taking the modulus of both sides and applying the triangle inequality, along with our hypothesis that $(a_{n})$ is decreasing, positive, we see that

$\begin{array}{lcl}\left|2i\sin\left(\frac{\theta}{2}\right)\right|\left|S_{N} - S_{M}\right| &=& \left|\sum_{j=M+2}^{N}\left(a_{j-1} - a_{j}\right)z^{j-\frac{1}{2}} - a_{M+1}z^{M+\frac{1}{2}} + a_{N}z^{N+\frac{1}{2}}\right|\\ &\leq&\sum_{j=M+2}^{N}\left|a_{j-1} - a_{j}\right|\left|z^{j-\frac{1}{2}}\right| + \left|a_{M+1}z^{M+\frac{1}{2}}\right| + \left|a_{N}z^{N+\frac{1}{2}}\right|\\&=&\sum_{j=M+2}^{N}\left(a_{j-1} - a_{j}\right) + a_{M+1} + a_{N}\\&=& a_{M+1} - a_{N} + a_{M+1} + a_{N}\\&=& 2a_{M+1}\end{array}$

Taking $M$ sufficiently large, we have that $2a_{M+1} < \epsilon$, for any $\epsilon > 0$ given, so the partial sums converge by the Cauchy criterion. $\Box$

Recall that a series is said to be Abel-summable if the limit

$\displaystyle \lim_{r \rightarrow 1^{-}}\sum_{k=0}^{\infty}a_{k}r^{k}$

exists.

The following theorem says that every convergent series (in the underlying topology) is Abel-summable. A classic example of a series in $\mathbb{R}$ which does not converge yet is Abel-summable is $\sum_{k=0}^{\infty}(-1)^{k}$. Indeed, for $0 \leq r < 1$, we have a geometric progression

$\displaystyle \sum_{k=0}^{N}(-1)^{k}r^{k} = \sum_{k=0}^{N}(-r)^{k} = \frac{1-(-r)^{N+1}}{1+r} \longrightarrow \frac{1}{2}$

as $r \rightarrow 1^{-}$.

Theorem 2. (Abel’s Summation Theorem)

Let $(a_{n})_{n=0}^{\infty}$ be a sequence of complex numbers such that the series

$\displaystyle F(z) := \sum_{k=0}^{\infty}a_{k}z^{k}$

converges for all $z \in \mathbb{D}$. If the series $\sum_{k=0}^{\infty}a_{k}$ converges, then

$\displaystyle \lim_{z \rightarrow 1^{-}}F(z) = \sum_{k=0}^{\infty}a_{k},$

where $z$ lies in a Stoltz angle $\left|1-z\right| \leq M(1-\left|z\right|)$ for some positive constant $M$.

Proof. Substracting a constant if necessary, we may assume without loss of generality that $\sum_{k=0}^{\infty}a_{k} = 0$. Fix a Stoltz angle $\left|1-z\right| \leq M(1-\left|z\right|)$. Setting $s_{n} := \sum_{k=0}^{n}a_{k}z^{k}$ and observing that $a_{n} = s_{n} - s_{n-1}$, we can rewrite $F(z)$ as

$\begin{array}{lcl} F(z) = a_{0} + \sum_{k=1}^{\infty}a_{k}z^{k} = a_{0} + \sum_{k=1}^{\infty}(s_{k} - s_{k-1})z^{k} &=& \sum_{k=0}^{\infty}s_{k}z^{k} - \sum_{k=0}^{\infty}s_{k}z^{k+1} \\&=& \left(1-z\right)\sum_{k=0}^{\infty}s_{k}z^{k}\end{array}$

Let $\epsilon > 0$ be given, and choose $N \in \mathbb{Z}^{\geq 1}$ sufficiently large so that $\left|s_{n}\right| < \epsilon$ for all $n \geq N$. By continuity, we can choose $\delta > 0$ such that

$\displaystyle\left|(1-z)\sum_{k=0}^{N}s_{k}z^{k}\right| < \epsilon$

For $z \in \mathbb{D}$ lying the Stoltz angle fixed above, we have the estimate

$\begin{array}{lcl} \left|\sum_{k=0}^{\infty}a_{k} - F(z)\right| = \left|F(z)\right| \leq \left|(1-z)\sum_{k=0}^{N}s_{k}z^{k}\right| + \left|(1-z)\sum_{k=N+1}^{\infty}s_{k}z^{k}\right| &<& \epsilon + \left|1-z\right|\sum_{k=N+1}^{\infty}\epsilon \left|z\right|^{k}\\&=& \epsilon + \epsilon\left|1-z\right|\frac{\left|z\right|^{N+1}}{1-\left|z\right|}\\&\leq& \epsilon + M(1-\left|z\right|)\frac{z^{N+1}}{1-\left|z\right|}\\&=& (M+1)\epsilon\end{array},$

which completes the proof. $\Box$

We remark that by scaling, Abel’s theorem generalizes to any series with radius of convergence $R> 0$.

Perhaps, the reader is still left with the question of why we care about Abel summability or what use is Abel’s theorem? First, the reader should note that given any bounded sequence of complex numbers $(a_{n})_{n=1}^{\infty}$, $F(z)$ is the generating function associated to $(a_{n})$. Abel’s theorem gives a method to compute closed-form expressions for infinite series. Before we give an example, we recall the alternating series test from calculus.

Lemma 3. (Alternating Series Test)

Let $(a_{k})_{k=0}^{\infty}$ be a bounded, decreasing sequence of nonnegative numbers such that $a_{k} \downarrow 0$. Then the series

$\displaystyle \sum_{k=0}^{\infty}(-1)^{k}a_{k}$

converges.

Proof. Fix $0 \in \mathbb{Z}^{\geq 0}$. Denote the $n^{th}$ partial sum by $s_{n}$. It is clear that $s_{2n+1} < s_{2n}$. Observe that

$\displaystyle s_{2n+2} = a_{0} + \sum_{k=0}^{n+1}\underbrace{(a_{2k} - a_{2k-1})}_{\leq 0} \leq s_{2n}, a_{0} \geq s_{2n+3} = \sum_{k=0}^{n}\underbrace{(a_{2k} - a_{2k+1})}_{\geq 0} \geq s_{2n+1}$

It follows from the monotone convergence theorem and the squeeze theorem that $s_{n} \rightarrow s \in \mathbb{R}$ and moreover,

$\displaystyle \left|s_{n} - s\right| \leq a_{n}$

$\Box$

Consider the sequence defined by $a_{k} := \frac{(-1)^{k}}{k+1}$, for $k \in \mathbb{Z}^{\geq 0}$. Then $\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}$ is convergent by the alternating series test, so

$\displaystyle F(z) := \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}z^{k}$

converges for $\left|z\right| < 1$. Since the geometric series convergence is uniform on a compact subsets of $\mathbb{D}$, we can compute $\int_{0}^{z}$, for $z \in (-1,0]$, term-by-term to obtain

$\displaystyle \ln(1+z) = \int_{-z}^{0}\frac{1}{1-u}du = \int_{-z}^{0}\left(\sum_{k=0}^{\infty}u^{k}\right)du = \sum_{k=0}^{\infty}\int_{-z}^{0}u^{k}du = \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}z^{k+1},$

which implies that $F(z) = \frac{\ln(1+z)}{z}$. Since $\frac{\ln(1+z)}{z} \rightarrow \ln(2)$ as $z \rightarrow 1^{-}$, we conclude from Abel’s theorem that

$\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1} = \ln(2)$