The Cauchy Condensation Test states that for a decreasing sequence where for all then converges if and only if converges.
Proof: In the forward direction, let us assume that converges. This means that the sequence of partial sums, converge. Define . We wish to show that the sequence converges.
Since for every it follows that and that . We may make similar arguments for the rest of the partial sum, resulting with
Thus for every . We are given that converges by our assumption and so any subsequence, specifically , also converges, and so also converges. So we have that is bounded above by a convergent sequence. Furthermore we know is increasing since for every . Thus we have a bounded, monotonic sequence and so it follows that converges. This implies that converges.
Now for the reverse direction. Assume that converges. Consider the partial sum
Since for all it follows that and . Continuing in this manner results in the following inequality:
I claim that this implies that converges. Let and consider . Since for all there is some such that . Thus the sequence is bounded and increasing. Every bounded monotonic sequence converges and so we may conclude that converges, thus, converges.
Reflection: This was another one of those problems where I was over-thinking. The proof really falls straight from definition, with some algebra tweaking along the way. The main thing I was worried about with this proof was the reverse direction, since I was able to bound a subsequence and not the whole sequence. The reason this wasn’t a problem in this case is because our are all greater than or equal to . So, I was able to bound every term in the sequence by my previously bounded terms in the subsequence. After stating that I felt a lot more comfortable concluding that the sequence converged.
This is a really nice theorem to know, especially since it’s and if and only if. Also, it’s given me another weapon in my arsenal with which to test for convergence.