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

1. $n a_n\rightarrow 0$ as $n\rightarrow\infty$ , and that
2. $f(r)$ converges to some finite $L$ as $r\rightarrow 1^-$ ;

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 Cesàro 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 () follows immediately. QED