Abel's Theorem

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

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

Proof
Let $\epsilon > 0$.

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

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

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

From Abel's Lemma, we have:


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

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

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

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

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

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


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

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

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

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

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:
 * $\displaystyle \left|{\sum_{k=0}^\infty a_k x^k - \sum_{k=0}^\infty a_k}\right| < \epsilon$

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

The result follows.