A question about Abel Discriminance in series I just read the proof of Abel discriminance in the series chapter. 
It says If ∑bn converges， and {an} is monotone with ｜an｜≤K， then series ∑anbn converges. in my mathematics analysis textbook.
But after reading  the proof for the discriminance, I just think the "Monotone" is useless. 
In the proof, let Bn=b1+b2+……+bn, and then we get the lemma:
If {an} is monotone, 丨Bi丨≤ K, then 丨∑aibi from 1 to m丨≤ K(丨a1丨 + 2丨am丨）.
The monotone thing is only used to prove that
∑丨ai-a(i-1)丨* 丨bn丨 is no greater than K*丨∑(ai-a(i-1))丨, which I think is obvious even without the monotone hypothesis.
So what on earth do we need the monotone hypothesis for? Is the Abel discriminance correct without the monotone hypothesis? Any suggestion will be appreciated.
 A: Without the monotone hypothesis, the theorem is false.
Take $b_n = (-1)^n/\sqrt{n}$  and $a_n = (-1)^n/\sqrt{n}$.  
We have $|a_n| \leqslant K = 1$ and convergence of the alternating series, 
$$\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} $$
However, 
$$\sum_{n=1}^\infty a_nb_n = \sum_{n=1}^\infty\frac{1}{n} = +\infty$$
How the proof applies the monotone hypothesis
Using summation by parts, the partial sum satisfies
$$\tag{*}\sum_{j=1}^n a_jb_j = B_na_n + \sum_{j=1}^{n-1}B_j(a_j - a_{j+1}),$$ 
where $B_n = \sum_{j=1}^n b_j$ and, by hypothesis, $\lim_{n \to \infty}B_n = B = \sum_{n=1}^\infty b_n$.
Since the sequence $(a_n)$ is monotone and bounded it must converge, with $\lim_{n \to \infty}a_n = a$. Thus, the first term on the RHS of (*) converges to $BA$.  This is the first place where the monotonicity of $(a_n)$ is used -- although for this part we could simply use the hypothesis that the sequence is convergent. 
To prove that the second term on the RHS of (*) is convergent we use that fact that $(B_n)$ is convergent and, hence, bounded with $|B_n| \leqslant M$ for some $M > 0$ and all $n \in \mathbb{N}.$  We also use the fact that $(a_n)$ is monotone and convergent.  However, the monotonicity is essential here. The counterexample shows that convergence of $(a_n)$ is not enough.
Note that $\sum_{j=1}^{n-1}|B_j(a_j - a_{j+1})| \leqslant M \sum_{j=1}^{n-1}|a_j - a_{j+1}|.$
If $(a_n)$ is monotone , then $|a_j - a_{j+1}| = a_j - a_{j+1}$  or   $|a_j - a_{j+1}| = -(a_j - a_{j+1})$ and, either, 
$$\sum_{j=1}^{n-1}|a_j - a_{j+1}| = \sum_{j=1}^{n-1}(a_j - a_{j+1}) = a_1 - a_n =|a_1 - a_n|,$$
or 
$$\sum_{j=1}^{n-1}|a_j - a_{j+1}| = -\sum_{j=1}^{n-1}(a_j - a_{j+1}) = a_n - a_1 = |a_1 - a_n|$$
Hence,
$$\sum_{j=1}^{\infty}|B_j(a_j - a_{j+1})| \leqslant M \sum_{j=1}^{\infty}|a_j - a_{j+1}| = M|a_1 - a|$$
Since this series is absolutely convergent, the second term on the RHS is convergent, which completes the proof.
