Abel's Theorem

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

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

where $\displaystyle \lim_{x \to 1^-}$ denotes the limit from the left.

Proof
Let $\epsilon > 0$.

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

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

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

From Abel's Lemma, we have:


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

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


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

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

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

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


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

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

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

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

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

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

The result follows.

Also known as
This result can also be found as Abel's lemma. However, the latter name is also found attached to a completely different result, so take care.