Abel's lemma is given in Abbott's Understanding Analysis as follows:
Let $b_n$ satisfy $b_1 \geq b_2 \geq …\geq 0$ and let $\sum_{n=1}^{\infty}a_n$ be a series for which the partial sums are bounded. In other words, assume there exists $A>0$ such that $|a_1+a_2+…+a_n| \leq A$ for all $n \in \mathbb N$. Then, for all $n \in \mathbb N$, $|a_1b_1+a_2b_2+…+a_nb_n| \leq 2Ab_1$.
The proof is not given but it is said that the proof follows from the summation by parts. I found here that the summation by parts is:
Consider two sequences $(a_n)$ and $(b_n)$. Let $S_N=\sum_{n=1}^{N}a_n$ be the n-th partial sum. Then for any $0 \leq m \leq n$ we have
$$\sum_{j=m}^na_jb_j=[S_nb_n-S_{m-1}b_m]+\sum_{j=m}^{n-1}S_j(b_j-b_{j+1})$$
Applying the above definition to Abel's lemma gives:
$$\sum_{j=1}^{n}a_jb_j=\sum_{j=1}^{n-1}(b_j-b_{j+1})S_j+S_nb_n-S_0b_1$$
But I am stuck here. I don't know the value of $S_0$ and I can't see a way to find $2Ab_1$ part of the Abel's lemma. Can you show me what to do?
Best Answer
First, $S_0 = 0$. Then, since $|S_j| \leqslant A$ and $b_j - b_{j+1} \geqslant 0,$ we can conclude
$$\left| \sum_{j=1}^{n-1} ( b_j - b_{j+1} ) S_j + S_n b_n \right| \leqslant |S_n| \cdot b_n + \sum_{j=1}^{n-1} ( b_j - b_{j+1} ) \cdot |S_j| \leqslant A \cdot b_n + A \cdot \sum_{j=1}^{n-1} ( b_j - b_{j+1} ).$$
Terms under sum will cancel out, giving:
$$A \cdot b_n + A \cdot \sum_{j=1}^{n-1} ( b_j - b_{j+1} ) = A \cdot \left[ b_n + ( b_1 - b_n)\right] = A b_1.$$
We actually acquired a stricter bound.