# proof of Tauber’s convergence theorem

Let

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

be a complex power series, convergent in the open disk $|z|<1$. We suppose that

1. 1.

$na_{n}\rightarrow 0$ as $n\rightarrow\infty$, and that

2. 2.

$f(r)$ converges to some finite $L$ as $r\rightarrow 1^{-}$;

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

 $\left|s_{n}-f\left(\frac{n-1}{n}\right)\right|\rightarrow 0\quad\text{as}\quad n% \rightarrow\infty.$ (1)

The desired result then follows in an obvious fashion.

For every real $0 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

 $\epsilon_{n}=\sup_{k>n}|ka_{k}|,$

and noting that

 $1-r^{k}=(1-r)(1+r+\cdots+r^{k-1})

we have that

 $|s_{n}-f(r)|\leq(1-r)\sum_{k=0}^{n}ka_{k}+\frac{\epsilon_{n}}{n}\sum_{k=n+1}^{% \infty}r^{k}.$

Setting $r=1-1/n$ in the above inequality we get

 $|s_{n}-f(1-1/n)|\leq\mu_{n}+\epsilon_{n}(1-1/n)^{n+1},$

where

 $\mu_{n}=\frac{1}{n}\sum_{k=0}^{n}|ka_{k}|$

are the Cesàro means of the sequence $|ka_{k}|,\;k=0,1,\ldots$ 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 (1) follows immediately. QED

Title proof of Tauber’s convergence theorem ProofOfTaubersConvergenceTheorem 2013-03-22 13:08:20 2013-03-22 13:08:20 rmilson (146) rmilson (146) 7 rmilson (146) Proof msc 40G10