Abel's Theorem

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Theorem

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


Then:

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

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 $\left \langle {s_N} \right \rangle$, 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: Formulation 2, 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:

\(\displaystyle \left\vert {\sum_{k \mathop = m}^n a_k x^k}\right\vert\) \(<\) \(\displaystyle \left({1 - x}\right) \sum_{k \mathop = m}^{n-1} \frac \epsilon 3 x^k + \frac \epsilon 3 x^n\) replacing instances of $\displaystyle \sum_{l \mathop = m}^k a_l$ with $\dfrac \epsilon 3$
\(\displaystyle \) \(<\) \(\displaystyle \frac \epsilon 3 \left({1 - x}\right) \frac {1 - x^n} {1 - x} + \frac \epsilon 3 x^n\) Sum of Geometric Progression
\(\displaystyle \) \(=\) \(\displaystyle \frac \epsilon 3\)


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.

$\blacksquare$


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