Abel's Test

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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 $K$.


Let b0 = 0, $B_N=\sum_{k=0}^N b_k$. Then bn = Bn − Bn − 1, n ≥ 1, hence \begin{align*} \sum_{k=1}^N a_kb_k &= \sum_{k=1}^N a_k(B_k-B_{k-1}) \\ &=B_1(a_1-a_2) +B_2(a_2-a_3)+\dots B_{N-1}(a_{N-1}-a_N)+a_NB_N\\ &=\sum_{k=1}^{N-1} B_k(a_k-a_{k+1})+a_NB_N \end{align*} [This identity is Abel's Lemma/Formulation 2.]

By Monotone Convergence Theorem {an} converges; and {BN} converges since bn converges. Hence the second addend aNBN converges.

It remains to prove the first addend Bk(akak+1) converges.

Since $\sequence{B_N}$ converges, $\bigsize{B_N}\le M$ for some $M$.

\(\ds \sum_1^N\bigsize{B_k (a_k-a_{k+1})}\) \(\le\) \(\ds M \sum_1^N \bigsize{a_k-a_{k+1} }\)
\(\ds \) \(=\) \(\ds M\bigsize{a_1-a_{N+1} }\) since {an} is decreasing
\(\ds \) \(\to\) \(\ds M \bigsize{a_1-a}\) since ak → a.

Hence Bk(ak − ak+1) converges absolutely.

Hence $\ds \sum a_n b_n$ converges.

Source of Name

This entry was named for Niels Henrik Abel.