Abel's Theorem

Theorem
Let $$\sum_{k=0}^\infty a_k$$ be a convergent series in $\R$.

Then $$\lim_{x \to 1^-} \left({\sum_{k=0}^\infty a_k x^k}\right) = \sum_{k=0}^\infty a_k$$.

Proof
Let $$\epsilon > 0$$.

Let $$\sum_{k=0}^\infty a_k$$ converge to $$s$$.

Then its sequence of partial sums $$\left \langle {s_N} \right \rangle$$, where $$s_N = \sum_{n=1}^N a_n$$, is a Cauchy sequence.

So $$\exists N: \forall k, m: k \ge m \ge N: \left|{\sum_{l=m}^k a_l}\right| < \frac \epsilon 3$$.

From Abel's Lemma, we have:

$$\sum_{k=m}^n u_k v_k = \sum_{k=m}^{n-1} \left({\left({\sum_{l=m}^k u_l}\right) \left({v_k - v_{k+1}}\right)}\right) + v_n \sum_{k=m}^n u_k$$.

We apply this, with $$u_k = a_k$$ and $$v_k = x^k$$:

$$\sum_{k=m}^n a_k x^k = \sum_{k=m}^{n-1} \left({\left({\sum_{l=m}^k a_l}\right) \left({x^k - x^{k+1}}\right)}\right) + x^n \sum_{k=m}^n a_k$$.

So it follows that $$\forall n \ge m \ge N$$ and $$\forall 0 < x < 1$$, we have:

$$ $$ $$

So we conclude that $$\left|{\sum_{k=N}^\infty a_k x^k}\right| \le \frac \epsilon 3$$.

Next, note that from the above, we have $$\forall x: 0 < x < 1$$:

$$\left|{\sum_{k=0}^\infty a_k x^k - \sum_{k=0}^\infty a_k}\right| \le \sum_{k=0}^{N-1} \left|{a_n}\right| \left({1 - x^n}\right) + \frac \epsilon 3 + \frac \epsilon 3$$.

But for finite $$n$$, we have that $$1 - x^n \to 0$$ as $$x \to 1^-$$.

Thus $$\sum_{k=0}^{N-1} \left|{a_n}\right| \left({1 - x^n}\right) \to 0$$ as $$x \to 1^-$$.

So $$\exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k=0}^{N-1} \left|{a_n}\right| \left({1 - x^n}\right) < \frac \epsilon 3$$.

So, for any given $$\epsilon > 0$$, we can find a $$\delta > 0$$ such that, for any $$x$$ such that $$1 - \delta < x < 1$$, it follows that $$\left|{\sum_{k=0}^\infty a_k x^k - \sum_{k=0}^\infty a_k}\right| < \epsilon$$.

That coincides with the definition for the limit from the left.

The result follows.