Abel's Lemma. Let $b_n$ satisfy $b_1\ge b_2\ge b_3\ge …\ge0$. Let $\sum _{n=1} ^\infty a_n$ be a series whose sequence of partial sums $s_m=\sum _{n=1} ^m a_n$ is bounded: there is $A \gt 0$ such that $\lvert s_m \rvert \le A$ for all $m \in \mathbb{N}$. Then for integers $n\gt m\ge0$ there holds $$\lvert \sum _{j=m+1} ^n a_jb_j \rvert \le 2Ab_{m+1}$$.
My proof:
Proof. Since $\lvert s_i \rvert \le A$ for all $i \in \mathbb{N}$, $\lvert s_p-s_q \rvert \le 2A$ for all $p,q\in\mathbb{N}$. Since $b_i\ge b_j$ if $i\le j$, $$\lvert \sum _{j=m+1} ^n a_jb_j \rvert\le\lvert \sum _{j=m+1} ^n a_jb_{m+1} \rvert=b_{m+1}\lvert \sum _{j=m+1} ^n a_j \rvert=b_{m+1}\lvert s_n-s_m\rvert\le2Ab_{m+1}$$, which is the inequality. $\square$
Textbook proof:
Lemma. $$\sum_{j=m+1}^na_jb_j=s_nb_{n+1}-s_mb_{m+1}+\sum_{j=m+1}^ns_j(b_j-b_{j+1})$$ holds for all $n\ge1$ and $m\ge0$.
Proof. (A bunch of proof, but the lemma got proved anyway)
Proof. By lemma, one have $$\lvert\sum_{j=m+1}^na_jb_j\rvert=\lvert s_nb_{n+1}-s_mb_{m+1}+\sum_{j=m+1}^ns_j(b_j-b_{j+1})\rvert$$ $$\le\lvert s_n\rvert b_{n+1}+\lvert s_m\rvert b_{m+1}+\sum_{j=m+1}^n\lvert s_j\rvert(b_j-b_{j+1})$$ $$\le Ab_{n+1}+Ab_{m+1}+A(b_{m+1}-b_{m+2})+A(b_{m+2}-b_{m+3})+…+A(b_n-b_{n+1})$$ $$=2Ab_{m+1}$$, which is the inequality. $\square$
My questions
The proof provided by the textbook is far more complex than my proof. So
1) Is my proof correct?
2) If yes, would there be some reasonable explanation about why the textbook chose the complex approach?
3) If no, what did I miss and how can I fix it?
Best Answer
Your proof is not correct. The wrong step is the first inequality:
This is correct only when $a_{j}>0$ for all $j=m+1,\ldots, n$, if $a_{j}<0$ for some $j$, then you got some terms cancel part of others.
So the next natural thought would be using the below firstly inequality to achieve your goal (to use the decreasing property of $(b_{j})$),$$\left|\sum_{j=m+1}^{n}a_{j}b_{j}\right|\leq \sum_{j=m+1}^{n}|a_{j}b_{j}|\leq b_{m+1}\sum_{j=m+1}^{n}|a_{j}|,$$ and the problem is reduced down to the evaluation of $|a_{j}|$. Using the similar trick you used third identity, we see that $$a_{j}=s_{j+1}-s_{j},$$ and thus $$\sum_{j=m+1}^{n}|a_{j}|=\sum_{j=m+1}^{n}|s_{j+1}-s_{j}|\leq\sum_{j=m+1}^{n}|s_{j+1}|+|s_{j}|\leq \sum_{j=m+1}^{n}2A=2(n-m)A.$$
Thus, if using your idea, then you get a larger (softer) upper bound $$\left|\sum_{j=m+1}^{n}a_{j}b_{j}\right|\leq 2(n-m)Ab_{m+1}.$$ This is because we enlarge the terms so many times that we enlarge to much.
Hence, the alternative way is what the book implies. You need start from re-writing the series so that you get a cancel out bunch of things using telescope sum and individually bound $|s_{m}|$ and $|s_{n}|$ so that you only have $2A$.