$\sum a_nb_n$ convergence theorem (Rudin)

Question

Why does the first inequality (in the chain) hold? Is $$\left\lvert \sum_{n=p}^{q-1}A_n(b_n - b_{n+1})\right\lvert \le M \left\lvert \sum_{n=p}^{q-1}(b_n - b_{n+1})\right\lvert$$ true? and why?

Answer 1

You missed an absolute value. However, consider $$ S:=\left|\sum_{p\le n< q}A_nc_n+A_qb_q+A_{p-1}(-b_{p-1})\right|, $$ where $c_n:=b_n-b_{n+1}$. Since $(b_n)$ is nonincreasing then $c_n\ge 0$ for all $n$. Moreover, since $-M \le A_n \le M$ for all $n$ you have that for each real $x$ it holds $$ \forall n,\,\,\,\,\,-M|x| \le A_n x \le M|x|. $$ Considering the above inequality for each term $A_nc_n$, $A_qb_q$ and $A_{p-1}(-b_{p-1})$, you conclude that $$ S\le M\left(\sum_{p\le n< q}c_n+b_q+b_{p-1}\right). $$