Consider the following statement:
Let $b_n$ satisfy $b_1 \ge b_2 \ge \dots \ge 0$ and let $\sum_{n=1}^\infty a_n$ be a series for which the partial sums are bounded i.e. there exists $A > 0$ such that $$ |a_1 + \dots + a_n|\le A$$
for all $n$. Then
$$ |a_1b_1 + \dots +a_n b_n| \le 2b_1 A$$
I think I can prove it like this:
$$ |a_1b_1 + \dots +a_n b_n| \le |a_1b_1 + a_2b_1 + \dots +a_n b_1| = b_1|a_1 + \dots + a_n| \le b_1 A \le 2 b_1 A$$
Why does this not work? The real proof needs summation by parts but I don't understand why.
Your proof is wrong consider $a_1=1,a_2=-1,a_3=1,.....$. Put$b_1=2,b_2=1,b_3=0,b_4=0,...$. Then the inequality$|a_1b_1 + \dots +a_n b_n| \le |a_1b_1 + a_2b_1 + \dots +a_n b_1| $ is violated. So the inequality which you thought would be true is actually false and hence the proof uses summation by parts