Rudin exercise 3.8 solution

Question

Problem . if the sequence $\sum\limits_{}{a_{n}}$ converges and $b_{n}$ is a monotonic bounded sequence then $\sum\limits_{} {a_{n}b_{n}}$ converges.

Solution. Let's use the identity: $\sum\limits_{n=p}^q {a_{n}b_{n}}$ = $\sum\limits_{n=p}^{q-1} {s_{n}(b_{n}-b_{n-1})}+s_{q}b_{q}-s_{p-1}b_{p}$, where $s_{n}=\sum\limits_{k=1}^n {a_{n}}$. Then $|\sum\limits_{n=p}^{q-1} {s_{n}(b_{n}-b_{n-1})}|<=|\sum\limits_{n=p}^{q-1} {s_{n}}|B$, where B is the difference of upper and lower bound of the sequence $b_{n}$. since $a_{n}$ is convergent we can make $|\sum\limits_{n=p}^{q-1} {s_{n}}|$ as small as we want and therefore we can make this $|\sum\limits_{n=p}^{q-1} {s_{n}(b_{n}-b_{n-1})}|$ as small as we want like $x/3$. Then let's take the absolute value of another term of the right side of the identity. $|s_{q}b_{q}|<=|s_{q}||b_{up}|$, where $b_{up}$ is the upper bound of the sequence and the same argument here too. we can make this term as small as we want. The same goes to the absolute value of the third term on the right side of the identity. Finally $|\sum\limits_{n=p}^{q} {a_{n}b_{n}}|<=|\sum\limits_{n=p}^{q-1} {s_{n}(b_{n}-b_{n-1})}|+|s_{q}b_{q}|+|s_{p-1}b_{p}|$. We can find $p$ and $q$ to make the right side small enough, completing the proof. Is this right? I am not satisfied what I have written here, I should have stated better in my opinion but the general idea of the proof is shown I suppose.

Answer 1

The general idea is correct. The expression “the upper bound of the sequence” is not appropriate (there are infinitely many upper bounds). I think that it would be better to say that $b_{up}$ is an upper bound of the sequence $\bigl(\lvert b_n\rvert\bigr)_{n\in\mathbb N}$. With this choice, $b_{up}$ is non-negative. Also, with this choice you have$$\left\lvert\sum_{n=p}^{q-1}s_n(b_n-b_{n-1})\right\rvert\leqslant2\left\lvert\sum_{n=p}^{q-1}s_n\right\rvert b_{up}.$$Again, the expression “the difference of upper and lower bound of the sequence $b_n$” is not appropriate.

Answer 2

Assume that $\{b_n\}$ is increasing. If $s_n=\sum_{k=1}^n a_k$, then $$ \sum_{k=1}^n a_kb_k=\sum_{k=1}^n (s_k-s_{k-1})b_k=\sum_{k=1}^n s_kb_k-\sum_{k=2}^n s_{k-1}b_k=s_nb_n+ \sum_{k=2}^{n}s_{k-1}(b_{k-1}-b_{k}). $$ Now $\{s_nb_n\}$ is convergent, as a product of convergent sequences, while the series $\sum s_{n-1}(b_{n-1}-b_{n})$ is dominated by the series $\sum M(b_{n}-b_{n-1})$, where $M=\sup_{n\in\mathbb N}|s_n|$, and since $\{b_{n}-b_{n-1}\}$ is absolutely summable, so is $\sum s_{n-1}(b_{n-1}-b_{n})$.