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 $b_n\to 0$?

  • $\begingroup$ 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. $\endgroup$ Mar 3, 2021 at 17:23
  • $\begingroup$ @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? $\endgroup$
    – Koro
    Mar 3, 2021 at 17:30
  • $\begingroup$ It's not Cauchy. The first step in your proof doesn't look right to me. $\endgroup$ Mar 3, 2021 at 17:35
  • $\begingroup$ @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. 😊 $\endgroup$
    – Koro
    Mar 3, 2021 at 17:42
  • $\begingroup$ @Vercassivelaunos: Please see the edit. $\endgroup$
    – Koro
    Mar 3, 2021 at 18:47

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Koro
    Mar 3, 2021 at 19:11
  • 1
    $\begingroup$ No, they wont have the opposite signs; it would be $a - (-a) = 2a$ so your upper bound there should be $M b_p - (- M b_q) = M (b_p + b_q)$ $\endgroup$
    – E-A
    Mar 3, 2021 at 19:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .