Rudin: Ch. 3 #8

Problem Statement: If \sum a_n converges and \{b_n\} is monotonic and bounded, prove that \sum a_nb_n converges.

Proof: Since we are given that \{b_n\} is bounded and monotonic we know there exists some B\in\mathbb{R} such that b_n\rightarrow B. Let \varepsilon>0, then there exists some N_1\in\mathbb{N} such that \left|b-n-B\right|<\varepsilon for every n>N_1. Note that this implies that for n>N_1 it follows that b_n<\varepsilon+B.

Furthermore, since \sum a_n converges there exists some N_2\in\mathbb{N} such that \left|\sum\limits_{n=k}^{m}a_n\right|<\dfrac{\varepsilon}{\left|\varepsilon+B\right|} whenever k,m>N_2.

Define N=max\{N_1,N_2\} and consider k,m>N. Then it follows that




Thus, \sum a_nb_n 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 a_n terms, i.e: they don’t all need to be positive.

What is really happening here is that the b_n 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 b_n terms must be within \varepsilon of the limit B. Another key thing to note is that the value to which \sum a_nb_n converges is not necessarily “nicely” related to B and \sum a_n. That is, \sum a_nb_n is not necessarily B\sum a_n.

This entry was posted in Analysis, Cauchy, Math, Sequence, Series. Bookmark the permalink.

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