Abel's Test

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Theorem

Let $\ds \sum b_n$ be a convergent real series.

Let $\sequence {a_n}$ be a decreasing sequence of positive real numbers.


Then the series $\ds \sum a_n b_n$ is also convergent.


Abel's Test for Uniform Convergence

Let $\sequence {\map {a_n} z}$ and $\sequence {\map {b_n} z}$ be sequences of complex functions on a compact set $K$.

Let $\sequence {\map {a_n} z}$ be such that:

$\sequence {\map {a_n} z}$ is bounded in $K$
$\ds \sum \size {\map {a_n} z - \map {a_{n + 1} } z}$ is convergent with a sum which is bounded in $K$
$\ds \sum \map {b_n} z$ is uniformly convergent in $K$.


Then $\ds \sum \map {a_n} z \map {b_n} z$ is uniformly convergent on $K$.


Proof



Let $b_0 = 0$.

Let $B_N = \ds \sum_{k \mathop = 0}^N b_k$.

Then:

$\forall n \ge 1: b_n = B_n − B_{n − 1}$

From Abel's Lemma:

$\ds \sum_{k \mathop = 1}^N a_k b_k = \sum_{k \mathop = 1}^{N - 1} B_k \paren {a_k - a_{k + 1} } + a_N B_N$


By the Monotone Convergence Theorem:

$\sequence {a_n}$ converges
$\sequence {B_N}$ converges since $\ds \sum b_n$ converges.

Hence the second addend $a_N B_N$ converges.

It remains to prove the first addend $\ds \sum B_k \paren {a_k − a_{k + 1} }$ converges.

Since $\sequence {B_N}$ converges, $\size {B_N} \le M$ for some $M \in \N$.

\(\ds \sum_1^N \size {B_k \paren {a_k -a_{k + 1} } }\) \(\le\) \(\ds M \sum_1^N \size {a_k - a_{k + 1} }\)
\(\ds \) \(=\) \(\ds M \size {a_1 - a_{N + 1} }\) as $\sequence {a_n}$ is decreasing
\(\ds \) \(\to\) \(\ds M \size {a_1 - a}\) as $a_k \to a$

Hence $\ds \sum B_k \paren {a_k − a_{k + 1} }$ converges absolutely.

Hence $\ds \sum a_n b_n$ converges.

$\blacksquare$


Source of Name

This entry was named for Niels Henrik Abel.


Sources