Suppose that $$\sum_{n=0}^\infty a_n=L$$ is a convergent series, and set $$f(r) = \sum_{n=0}^\infty a_n r^n.$$ Convergence of the first series implies that $a_n\rightarrow 0$ , and hence $f(r)$ converges for $|r|<1$ . We will show that $f(r)\rightarrow L$ as $r\rightarrow 1^-$ .

Let $$s_N=a_0+\cdots+a_N,\quad N\in\natnums,$$ denote the corresponding partial sums. Our proof relies on the following identity \begin{equation} \label{eq:ident} f(r)=\sum_n a_n r^n = (1-r) \sum_n s_n r^n. \end{equation}The above identity obviously works at the level of formal power series. Indeed,

Let $\epsilon>0$ be given. Choose an $N$ sufficiently large so that all partial sums, $s_n$ with $n>N$ , satisfy $|s_n-L|\le\epsilon$ . Then, for all $r$ such that $0<r<1$ , one obtains $$\left|\sum_{n=N+1}^\infty (s_n - L) r^n\right| \le \epsilon\,\frac{r^{N+1}}{1-r}\,.$$ Note that $$f(r) - L = (1-r)\sum_{n=0}^N (s_n - L) r^n + (1-r) \sum_{n=N+1}^\infty (s_n - L) r^n.$$ As $r\rightarrow 1^-$ , the first term tends to $0$ . The absolute value of the second term is estimated by $\epsilon\, r^{N+1}\le \epsilon$ . Hence, $$\limsup_{r\rightarrow 1^-} |f(r) - L| \le \epsilon.$$ Since $\epsilon>0$ was arbitrary, it follows that $f(r)\rightarrow L$ as $r\rightarrow 1^-$ . QED