Abel's Limit Theorem

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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 1

Let $\epsilon > 0$.

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

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

So:

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

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

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


Let $u_k = a_k$ and $v_k = x^k$.

We have:

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


So $\forall n \in \N: n \ge m \ge N$ and $\forall x \in \R: 0 < x < 1$, 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 $\ds \sum_{l \mathop = m}^k a_l$ 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
\(\ds \) \(=\) \(\ds \frac \epsilon 3\)


We conclude that:

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


From the above, we have $\forall x \in \R: 0 < x < 1$:

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


But for finite $n$, we have:

$1 - x^n \to 0$ as $x \to 1^-$.

Thus:

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

So:

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

So, for any given $\epsilon > 0$, there exists $\delta > 0$, such that:

$\forall x \in \R: 1 - \delta < x < 1 \implies \ds \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$

The result follows from the definition for the limit from the left.

$\blacksquare$


Proof 2


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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.


Examples

Arbitrary Example 1

Let $\ds \map g x = \sum_{n \mathop \ge 1} \paren {-1}^{n - 1} \dfrac {x^n} n$ for $\size x < 1$.

Then:

$\map g x = \map \ln {1 + x}$

for $\size x < 1$.


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The series $\map g 1$ converges by Alternating Series Test,


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so by Abel's Limit Theorem:

$\map g 1 = \ds \lim _{x \mathop \to 1^{-} } \map g x = \lim_{x \mathop \to 1^{-} } \map \ln {1 + x} = \ln 2$

since the logarithm is a continuous function.


Arbitrary Example 2


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Let:

$\ds \map g x = \sum_{n \mathop \ge 0} \frac {\paren {-1}^{n - 1} \paren {2 } !} {2^{2 n} n!^2 \paren {2 n - 1} } x^n$

for $\size x < 1$.

Then:

$\map g x = \sqrt {1 + x}$

for $\size x < 1$.


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The series $\map g 1$ is absolutely convergent


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so by Abel's Limit Theorem and the continuity of $\sqrt {1 + x}$:

$\map g 1 = \ds \lim_{x \mathop \to 1^{-} } \map g x = \lim_{x mathop \to 1^{-} } \sqrt {1 + x} = \sqrt 2$


Arbitrary Example 3


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Let:

$\map g x = \dfrac 1 {1 + x^2}$

which is differentiable for all real $x$.


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When $\size x < 1$, $\map g x = \ds \sum_{n \mathop \ge 0} \paren {-1}^n x^{2 n}$ by expanding a geometric series.


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While $\map g x$ has a limit as $x \to 1^{-}$ (namely $1/2$), the power series does not converge at $x = 1$.


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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.


Sources

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