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
$$\left{s}_{n}f\left(\frac{n1}{n}\right)\right\to 0\mathit{\hspace{1em}}\text{as}\mathit{\hspace{1em}}n\to \mathrm{\infty}.$$  (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}.$$ 
Setting
$${\u03f5}_{n}=\underset{k>n}{sup}k{a}_{k},$$ 
and noting that
$$ 
we have that
$${s}_{n}f(r)\le (1r)\sum _{k=0}^{n}k{a}_{k}+\frac{{\u03f5}_{n}}{n}\sum _{k=n+1}^{\mathrm{\infty}}{r}^{k}.$$ 
Setting $r=11/n$ in the above inequality^{} we get
$${s}_{n}f(11/n)\le {\mu}_{n}+{\u03f5}_{n}{(11/n)}^{n+1},$$ 
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 ${\u03f5}_{n}$. Finally, Euler’s formula for $e$ gives
$$\underset{n\to \mathrm{\infty}}{lim}{(11/n)}^{n}={e}^{1}.$$ 
The validity of (1) follows immediately. QED
Title  proof of Tauber’s convergence theorem 

Canonical name  ProofOfTaubersConvergenceTheorem 
Date of creation  20130322 13:08:20 
Last modified on  20130322 13:08:20 
Owner  rmilson (146) 
Last modified by  rmilson (146) 
Numerical id  7 
Author  rmilson (146) 
Entry type  Proof 
Classification  msc 40G10 