proof of Tauber’s convergence theorem

Let

$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n,$$

be a complex power series, convergent in the open disk . We suppose that

1. $n{a}_n\to 0$ as $n\to \mathrm{\infty }$, and that

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

and wish to show that ${\sum }_n{a}_n$ converges to the same $L$ as well.

Let $s_n=a_0+\mathrm{\cdots }+a_n$, where $n=0,1,\mathrm{\dots }$, 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 $n\to \mathrm{\infty }$, and the evolution of $f(r)$ as $r\to 1^{-}$. Indeed we shall show that

$$\left|s_n-f\left(\frac{n-1}{n}\right)\right|\to 0\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}n\to \mathrm{\infty }.$$ (1)

The desired result then follows in an obvious fashion.

For every real we have

$$s_n=f(r)+\sum _{k=0}^n{a}_k(1-r^k)-\sum _{k=n+1}^{\mathrm{\infty }}{a}_k{r}^k.$$

Setting

$${ϵ}_n=\underset{k>n}{sup}|k{a}_k|,$$

and noting that

we have that

$$|s_n-f(r)|\le (1-r)\sum _{k=0}^{n}k{a}_k+\frac{{ϵ}_n}{n}\sum _{k=n+1}^{\mathrm{\infty }}{r}^k.$$

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

$$|s_n-f(1-1/n)|\le {\mu }_n+{ϵ}_n{(1-1/n)}^{n+1},$$

where

$${\mu }_n=\frac{1}{n}\sum _{k=0}^n|k{a}_k|$$

are the Cesàro means of the sequence $|k{a}_k|,k=0,1,\mathrm{\dots }$ Since the latter sequence converges to zero, so do the means ${\mu }_n$, and the suprema ${ϵ}_n$. Finally, Euler’s formula for $e$ gives

$$\underset{n\to \mathrm{\infty }}{lim}{(1-1/n)}^n=e^{-1}.$$

The validity of (1) follows immediately. QED

Title	proof of Tauber’s convergence theorem
Canonical name	ProofOfTaubersConvergenceTheorem
Date of creation	2013-03-22 13:08:20
Last modified on	2013-03-22 13:08:20
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	7
Author	rmilson (146)
Entry type	Proof
Classification	msc 40G10