ProofOfTaubersConvergenceTheorem 2002-11-07 16:09:12 2008-08-07 10:30:55 Proof Abel summability 0 \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \newcommand{\reals}{\mathbb{R}} \newcommand{\natnums}{\mathbb{N}} \newcommand{\cnums}{\mathbb{C}} \newcommand{\znums}{\mathbb{Z}} \newcommand{\lp}{\left(} \newcommand{\rp}{\right)} \newcommand{\lb}{\left[} \newcommand{\rb}{\right]} \newcommand{\supth}{^{\text{th}}} \newtheorem{proposition}{Proposition} \newtheorem{definition}[proposition]{Definition} \newtheorem{theorem}[proposition]{Theorem} Let $$f(z) = \sum_{n=0}^\infty a_n z^n,$$ be a complex power series, convergent in the open disk $\vert z\vert<1$. We suppose that \begin{enumerate} \item $n a_n\rightarrow 0$ as $n\rightarrow\infty$, and that \item $f(r)$ converges to some finite $L$ as $r\rightarrow 1^-$; \end{enumerate} and wish to show that $\sum_n a_n$ converges to the same $L$ as well. Let $s_n=a_0+\cdots+a_n$, where $n=0,1,\ldots$, denote the partial sums of the series in question. The enabling idea in Tauber's convergence result (as well as other Tauberian theorems) is the existence of a correspondence in the evolution of the $s_n$ as $n\rightarrow\infty$, and the evolution of $f(r)$ as $r\rightarrow 1^-$. Indeed we shall show that \begin{equation} \label{eq:e1} \left\vert s_n - f\lp \frac{n-1}{n} \rp\right\vert \rightarrow 0 \quad\text{as}\quad n\rightarrow \infty. \end{equation} The desired result then follows in an obvious fashion. For every real $0<r<1$ we have $$s_n = f(r) + \sum_{k=0}^n a_k (1-r^k) - \sum_{k=n+1}^\infty a_k\, r^k.$$ Setting $$\epsilon_n = \sup_{k>n} \vert k a_k \vert,$$ and noting that $$ 1-r^k = (1-r)(1+r+\cdots+r^{k-1}) < k(1-r),$$ we have that $$ \vert s_n - f(r) \vert \leq (1-r)\sum_{k=0}^n ka_k + \frac{\epsilon_n}{n} \sum_{k=n+1}^\infty r^k.$$ Setting $r=1-1/n$ in the above inequality we get $$ \vert s_n - f(1-1/n) \vert \leq \mu_n + \epsilon_n (1-1/n)^{n+1},$$ where $$\mu_n = \frac{1}{n}\sum_{k=0}^n \vert k a_k \vert$$ are the Ces\`aro means of the sequence $\vert k a_k\vert,\; k=0,1,\ldots$ Since the latter sequence converges to zero, so do the means $\mu_n$, and the suprema $\epsilon_n$. Finally, Euler's formula for $e$ gives $$\lim_{n\rightarrow\infty}(1-1/n)^n = e^{-1}.$$ The validity of \eqref{eq:e1} follows immediately. QED