Proof of Abel's theorem by summation by parts

Proof

We rewrite $G$ with the partial sums by summation by parts.

1. Let $s_n = \sum_{k=0}^{n} a_k$ be the partial sums, and let $s = \lim s_n$, which exists by hypothesis. Put $s_{-1} = 0$, so that $a_n = s_n - s_{n-1}$ for all $n \ge 0$.
2. The sequence $(s_n)$ converges, so it is bounded: there is $K$ with $\abs{s_n} \le K$ for all $n$. Then $\abs{a_n} \le 2K$ by the triangle inequality.
3. Fix $x$ with $0 \le x < 1$. Then $\abs{a_n x^n} \le 2K x^n$ and $\abs{s_n x^n} \le K x^n$. The geometric series $\sum x^n$ converges.
4. By the comparison test, $\sum a_n x^n$ and $\sum s_n x^n$ converge absolutely. So both converge.
5. For each $N$, $$\sum_{n=0}^{N} a_n x^n = \sum_{n=0}^{N} s_n x^n - \sum_{n=1}^{N} s_{n-1} x^n = (1 - x) \sum_{n=0}^{N} s_n x^n + s_N x^{N+1}.$$
6. We have $\abs{s_N x^{N+1}} \le K x^{N+1}$, and $x^{N+1} \to 0$. So $s_N x^{N+1} \to 0$.
7. Let $N \to \infty$ in step 5. The sum rule and the product rule for limits give $G(x) = (1 - x) \sum_{n=0}^{\infty} s_n x^n$.
8. The geometric series gives $(1 - x) \sum_{n=0}^{\infty} x^n = 1$. So $s = (1 - x) \sum_{n=0}^{\infty} s x^n$.
9. Subtract step 8 from step 7, using the limit rules on partial sums: $G(x) - s = (1 - x) \sum_{n=0}^{\infty} (s_n - s) x^n$.
10. Let $\varepsilon > 0$. By the epsilon-N form, there is $N$ with $\abs{s_n - s} < \varepsilon / 2$ for all $n \ge N$. Let $C = \sum_{n=0}^{N-1} \abs{s_n - s}$.
11. Split the series of step 9 into the terms with $n < N$ and the tail with $n \ge N$. Since $x^n \le 1$, the first part has absolute value at most $C$.
12. The tail has terms of absolute value at most $(\varepsilon/2) x^n$. By the bound for absolutely convergent series and the comparison test, its absolute value is at most $(\varepsilon/2) \sum_{n \ge N} x^n$.
13. The geometric series gives $(1 - x) \sum_{n \ge N} x^n = x^N \le 1$. With steps 9, 11 and 12, $\abs{G(x) - s} \le (1 - x) C + \varepsilon/2$.
14. Let $\delta = \min(1, \varepsilon / (2C + 1))$. If $1 - \delta < x < 1$, then $0 \le x < 1$, and $(1 - x) C \le \delta C \le C \varepsilon / (2C + 1) < \varepsilon / 2$.
15. So $\abs{G(x) - s} < \varepsilon$ for all $x$ with $1 - \delta < x < 1$. By the definition of the left limit, $G(x) \to s = \sum_{n=0}^{\infty} a_n$ as $x \to 1$ from the left.