# Abel's Theorem

## Theorem

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

Then:

- $\displaystyle \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$

where $\displaystyle \lim_{x \mathop \to 1^-}$ 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 $\sequence {s_N}$, where $\displaystyle s_N = \sum_{n \mathop = 1}^N a_n$, is a Cauchy sequence.

So:

- $\displaystyle \exists N: \forall k, m: k \ge m \ge N: \size {\sum_{l \mathop = m}^k a_l} < \frac \epsilon 3$

From Abel's Lemma: Formulation 2, we have:

- $\displaystyle \sum_{k \mathop = m}^n u_k v_k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k u_l} \paren {v_k - v_{k+1}} } + v_n \sum_{k \mathop = m}^n u_k$

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

- $\displaystyle \sum_{k \mathop = m}^n a_k x^k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k a_l} \paren {x^k - x^{k+1}} } + x^n \sum_{k \mathop = m}^n a_k$

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

\(\ds \size {\sum_{k \mathop = m}^n a_k x^k}\) | \(<\) | \(\ds \paren {1 - x} \sum_{k \mathop = m}^{n-1} \frac \epsilon 3 x^k + \frac \epsilon 3 x^n\) | replacing instances of $\displaystyle \sum_{l \mathop = m}^k a_l$ with $\dfrac \epsilon 3$ | |||||||||||

\(\ds \) | \(<\) | \(\ds \frac \epsilon 3 \paren {1 - x} \frac {1 - x^n} {1 - x} + \frac \epsilon 3 x^n\) | Sum of Geometric Progression | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac \epsilon 3\) |

So we conclude that:

- $\displaystyle \size {\sum_{k \mathop = N}^\infty a_k x^k} \le \frac \epsilon 3$

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

- $\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} \le \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} + \frac \epsilon 3 + \frac \epsilon 3$

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

Thus:

- $\displaystyle \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} \to 0$ as $x \to 1^-$

So:

- $\displaystyle \exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} < \frac \epsilon 3$

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:

- $\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$

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

The result follows.

$\blacksquare$

## 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.

## Source of Name

This entry was named for Niels Henrik Abel.

## Historical Note

Abel's Theorem first appears in Abel's paper *Mémoire sur une Propriété Générale d'une Classe Très-Étendue de Fonctions Transcendantes*.

Carl Gustav Jacob Jacobi remarked that it was the greatest discovery in integral calculus in the $19$th century.

## Sources

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous): Appendix: $\S 18.9$: Continuity and differentiation of power series