## 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

$\left|\sum\limits_{n=k}^{m}a_nb_n\right|<\left|\sum\limits_{n=k}^{m}a_n(\varepsilon+B)\right|$

$=\left|\varepsilon+B\right|\left|\sum\limits_{n=k}^{m}a_n\right|$

$<\left|\varepsilon+B\right|\dfrac{\varepsilon}{\left|\varepsilon+B\right|}=\varepsilon$

Thus, $\sum a_nb_n$ is convergent.

$\Box$

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 $B\sum a_n$.