# Conditions given in theorem 3.42 in Rudin's Principal of Mathematical Analysis

The theorem states that if partial sums $$A_n$$ of series $$\sum a_n$$ are bounded and $$(b_n)$$ is a monotonically decreasing sequence such that $$b_n\to 0$$, then the series $$\sum a_nb_n$$ converges.

I tried to prove it as follows:
Since $$A_n$$'s are bounded, there exists $$M\gt 0$$ such that $$|A_n|\lt M$$ for all $$n\in \mathbb N$$. Since $$(b_n)$$ converges, for every $$\epsilon\gt 0, \exists N:\forall q\ge p\ge N,$$ we have $$b_p-b_q\lt \frac \epsilon {2M}$$.

We have, $$|\sum_{n=p}^q a_nb_n|=|\sum_{n=p}^{q-1}A_n(b_n-b_{n+1})+A_qb_q-A_{p-1}b_p|$$ $$\lt M |\sum_{n=p}^{q-1}(b_n-b_{n+1})+b_p-b_q|=2M(b_p-b_q)\lt \epsilon \tag 1$$
Hence, by Cauchy Criterion, the series $$\sum a_nb_n$$ converges.

Please note that in the above proof, I have nowhere used $$b_n\to 0$$. I have only used the fact that the sequence $$(b_n)$$ is monotonically decreasing and convergent.

My question is why $$b_n$$ must converge to $$0$$? Clearly, from the proof above, convergence of $$(b_n)$$ is enough, irrespective of to what $$\alpha\in \mathbb R$$, $$(b_n)$$ converges.

Edit: Suppose that $$(b_n)$$ is a sequence of non-negative nos. We have $$-M\lt A_q\lt M\implies -Mb_q\lt A_qb_q\lt Mb_q$$. Similarly, $$-Mb_p\lt A_{p-1}b_p\lt Mb_p$$. It follows that $$-M(b_p-b_q)\lt A_{p-1}b_p-A_qb_q\lt M(b_p-b_q)\implies |A_{p-1}b_p-A_qb_q|\lt M(b_p-b_q)$$ so $$(1)$$ holds. Is it necessary for $$b_n\to 0$$?

• If $b_n$ does not converge to $0$, then $a_nb_n$ may also not converge to $0$, for instance, if $a_n=(-1)^n$. By the trivial criterion, the series does not converge in this case. Your first step already looks iffy to me. Mar 3, 2021 at 17:23
• @Vercassivelaunos:I thought along the same lines. If $b_n$ converges but not to $0$ then there exists some $t\gt 0$ such that $b_n\gt t$ for infinitely many $n$ and taking $a_n$ as you have taken. $a_{k_{2n}}b_{k_{2n}}= (A_{k_{2n}}-A_{{k_{2n}+1}})b_{k_{2n}}\gt 2t \nrightarrow 0$. Hence, the series is divergent, but then why is it Cauchy?
– Koro
Mar 3, 2021 at 17:30
• It's not Cauchy. The first step in your proof doesn't look right to me. Mar 3, 2021 at 17:35
• @Vercassivelaunos; I think that you are right. I had unconsciously assumed $(b_n)$ to be sequence of non negative terms. That's why I wrote the first step. Thanks a lot. ðŸ˜Š
– Koro
Mar 3, 2021 at 17:42
• @Vercassivelaunos: Please see the edit.
– Koro
Mar 3, 2021 at 18:47

Your edit clarifies the issue in your proof; indeed, if that proof were valid, why, any bounded set of numbers would have to be equal to one another! (set $$b_p = b_q = 1$$)
The issue is how you are subtracting the inequalities: If $$a_l < a < a_u$$ and $$b_l < b < b_u$$, then $$a_l - b_u < a - b < a_u - b_l$$ and not $$a_l - b_l < a - b < a_l - b_l$$. Intuitively, the largest $$a - b$$ can be is if $$a$$ is its largest, and $$b$$ is its smallest value.
• Thanks for response. I understand that if $a_l < a < a_u$ and $b_l < b < b_u$ then $a_l - b_u < a - b < a_u - b_l$. But that's not the issue in my case because if $-a\lt b\lt a$ then $-a\lt -b\lt a$. What I am trying to convey is both sides of my inequality have the same no. with opposite signs.
• No, they wont have the opposite signs; it would be $a - (-a) = 2a$ so your upper bound there should be $M b_p - (- M b_q) = M (b_p + b_q)$