根据 $A_n$ 的定义,
\[
\begin{aligned}
A_n&=a_{N+1}+a_{N+2}+\cdots+a_{n-1}+a_n,\\
A_{n-1}&=a_{N+1}+a_{N+2}+\cdots+a_{n-1},\\
\end{aligned}
\]
因此, $a_n=A_n-A_{n-1}$, 这里 $N < n\leqslant N+M$. 于是,
\[
\begin{split}
\sum_{N < n\leqslant N+M}a_n b_n &=\sum_{N < n\leqslant N+M}(A_n-A_{n-1})b_n\\
&=\sum_{N < n\leqslant N+M}A_n b_n-\sum_{N < n\leqslant N+M}A_{n-1} b_n\\
&=\sum_{N < n\leqslant N+M}A_n b_n-\sum_{N-1 < n\leqslant N+M-1}A_{n} b_{n+1}\\
&=\sum_{N < n\leqslant N+M}A_n b_n-\Biggl(\sum_{N < n\leqslant N+M}A_{n} b_{n+1}+A_{N} b_{N+1}-A_{N+M}b_{N+M+1}\Biggr)\\
&=\sum_{N < n\leqslant N+M}A_n (b_n-b_{n+1})-A_{N} b_{N+1}+A_{N+M}b_{N+M+1}
\end{split}
\]
这里要注意到 $A_N$ 可以认为是 0. 所以
\[
\sum_{N < n\leqslant N+M}a_n b_n =A_{N+M}b_{N+M+1}+\sum_{N < n\leqslant N+M}A_n (b_n-b_{n+1}).
\]
现在 $b_n-b_{n+1}\geqslant 0$, $\forall\ n > N$. 于是由上面的不等式, 可得
\[
\begin{split}
\biggl|\sum_{N < n\leqslant N+M}a_n b_n \biggr|&\leqslant Ab_{N+M+1}+A\sum_{N < n\leqslant N+M}(b_n-b_{n+1})\\
&=Ab_{N+M+1}+A\Big[(b_{N+1}-b_{N+2})+(b_{N+2}-b_{N+3})+\cdots+(b_{N+M}-b_{N+M+1})\Bigr]\\
&=Ab_{N+1}.
\end{split}
\]