The Unapologetic Mathematician

Mathematics for the interested outsider

Dirichlet’s and Abel’s Tests

We can now use Abel’s partial summation formula to establish a couple other convergence tests.

If a_n is a sequence whose sequence A_n of partial sums form a bounded sequence, and if B_n is a decreasing sequence converging to zero, then the series \sum_{k=0}^\infty a_kB_k converges. Indeed, then the sequence A_nB_{n+1} also decreases to zero, so we just need to consider the series \sum_{k=0}^\infty A_kb_{k+1}.

The bound on |A_k| and the fact that B_k is decreasing imply that |A_k(B_k-B_{k+1})|\leq M(B_k-B_{k+1}), and the series \sum_{k=0}^\infty M(B_k-B_{k+1}) clearly converges. Thus by the comparison test, the series \sum_{k=0}^\infty A_kb_{k+1} converges absolutely, and our result follows. This is called Dirichlet’s test for convergence.

Let’s impose a bit more of a restriction on the A_n and insist that this sequence actually converge. Correspondingly, we can weaken our restriction on B_n and require that it be monotonic and convergent, but not specifically decreasing to zero. These two changes balance out and we still find that \sum_{k=0}^na_kB_k converges. Indeed, the sequence A_nB_{n+1} converges automatically as the product of two convergent sequences, and the rest is similar to the proof in Dirichlet’s test. We call this Abel’s test for convergence.

May 1, 2008 - Posted by | Analysis, Calculus


  1. Think you have a mistake:

    If B_k is decreasing, then:

    B_{k+1} – B_k < 0.

    So it is not the case that:

    |A_k(B_{k+1} – B_k)| <= M(B_{k+1} – B_k)

    Comment by David | November 27, 2011 | Reply

  2. Sorry, yes, the indices are swapped.

    Comment by John Armstrong | November 28, 2011 | Reply

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