Problem Statement: If converges and is monotonic and bounded, prove that converges.
Proof: Since we are given that is bounded and monotonic we know there exists some such that . Let , then there exists some such that for every . Note that this implies that for it follows that .
Furthermore, since converges there exists some such that whenever .
Define and consider . Then it follows that
Thus, is convergent.
Reflection: When I first attacked this question I tried looking at the partial sums alone, not using Cauchy, and I got to a point where my bound depended on which partial sum I was at. Bad news bears! So, I scratched that idea and considered a new approach. Once I started playing with Cauchy it sort of fell out on it’s own. The main idea with this problem is that you can take any convergent series and multiply each term by a monotonic and bounded sequence and the resulting series will still be convergent. This is really handy because there are no restrictions about terms, i.e: they don’t all need to be positive.
What is really happening here is that the terms of the sequence can do funny things to the terms of the series, but they can only do strange and crazy things to finitely many terms of the series since eventually the terms must be within of the limit . Another key thing to note is that the value to which converges is not necessarily “nicely” related to and . That is, is not necessarily .