# 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