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 | 2 Comments



Get every new post delivered to your Inbox.

Join 452 other followers