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 partialsums of the series in question. The enabling idea in Tauber’sconvergence result (as well as other Tauberian theorems) is theexistence 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 Cesàromeans of the sequence $|k{a}_k|,k=0,1,\mathrm{\dots }$Since thelatter sequence converges to zero, so do the means ${\mu }_n$, and thesuprema ${ϵ}_n$. 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