Abel's Limit Theorem/Proof 2
Theorem
Let $\ds \sum_{k \mathop = 0}^\infty a_k$ be a convergent series in $\R$.
Then:
- $\ds \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
where $\ds \lim_{x \mathop \to 1^-}$ denotes the limit from the left.
Proof
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Since $\ds \sum_{k \mathop = 0}^\infty a_k$ converges and $\cmod{x^k}\le1$ and $\sequence{x^k}$ is decreasing, by Abel's Test for Uniform Convergence $\ds \sum_{k \mathop = 0}^\infty a_kx^k$ converges uniformly on $[0,1]$ and hence to a continuous function by Uniform Limit Theorem.
Also known as
Abel's Limit Theorem is also known just as Abel's Theorem.
However, the latter name has more than one theorem attached to it, so the full name is preferred.
Again, Abel's Limit Theorem can also be found as Abel's Lemma.
However, the latter name is also found attached to a completely different result, so again, it is preferred that it not be used in this context.
Also see
Source of Name
This entry was named for Niels Henrik Abel.