This is from Rudin PMA:

Theorem 3.42 Suppose

(a) the partial sums $A_n$ of $\sum a_n$ form a bounded sequence;

(b) $b_0\geqslant b_1\geqslant b_2\geqslant \dots;$

(c) $\lim_{n\to \infty} b_n=0.$

Then $\sum a_nb_n$ converges.

The proof goes like this:

Choose $M$ such that $|A_n|\le M$ for all $n$. Given $\varepsilon>0$, there is an integer $N$ such that $b_n\le\dfrac{\varepsilon}{2M}$. For $N\le p\le q$, we have $$\left|\sum_{n=p}^q a_nb_n\right|=\left|\sum_{n=p}^{q-1} A_n(b_n-b_{n+1})+A_qb_q-A_{p-1}b_p\right|\le M\left|\sum_{n=p}^{q-1} (b_n-b_{n+1})+b_q+b_p\right|$$.

I can't understand how this last inequality is derived. Rudin claims that it depends on the fact that $b_n-b_{n+1}\ge0$.


1 Answer 1


\begin{align}\left|\sum_{n=p}^{q-1} A_n(b_n-b_{n+1})+A_qb_q-A_{p-1}b_p\right|&\leqslant\sum_{n=p}^{q-1}|A_n|(b_n-b_{n+1})+|A_q|b_q+|A_{p-1}|b_p\\&\leqslant M\left(\sum_{n=p}^{q-1}(b_n-b_{n+1})+b_q+b_p\right).\end{align}

  • $\begingroup$ If $(b_n)$ is a sequence of non negative nos., then is true that convergence of $(b_n)$ is enough i.e., it's not necessary that $b_n\to 0$? $\endgroup$
    – Koro
    Mar 3, 2021 at 19:02
  • $\begingroup$ No, the convergence of $(b_n)_{n\in\Bbb N}$ is not enough. $\endgroup$ Mar 3, 2021 at 19:03
  • $\begingroup$ Thanks a lot sir. I tried to prove this here math.stackexchange.com/questions/4047392/…. May I request you to please have a look at it and help me understand what is wrong in my proof? Thanks. $\endgroup$
    – Koro
    Mar 3, 2021 at 19:05

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