# The Unapologetic Mathematician

## 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 »

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