# Abel criterion proof

I want to prove the following statement:

Let be $$\sum\limits_{k=1}^{\infty} a_k$$ a convergent series and $$\left(b_k\right)_{n\in\mathbb{N}}$$ a monotone and bounded sequence. Then $$\sum\limits_{k=1}^{\infty} a_kb_k$$ is also convergent.

I know there already exist a few questions on this problem, however they mostly have additional assumptions (i.e. $$\left(b_k\right)_{k\in\mathbb{N}}$$ with $$b_k\geq 0$$ for all $$k$$).

My approach:

We define $$A_n:=\sum\limits_{k=1}^{n} a_k$$. As $$A_n$$ is convergent there exists a bound $$A$$ such that $$|A_n|\leq A$$ for all $$n$$. We know that $$\left(b_k\right)_{k\in\mathbb{N}}$$ is convergent and hence the sequence $$\left(A_kb_k\right)_{k\in\mathbb{N}}$$ is also convergent (product of two convergent sequences). Let be $$n_1$$ and $$n_2$$ two indices such that for all $$n,m$$ with $$n>m>n_1$$ it holds $$|A_nb_n-A_mb_m|<\frac{\epsilon}{2}$$ and for all $$n,m$$ with $$n>m>n_2$$ it holds $$|b_n-b_m|<\frac{\epsilon}{2A}$$. Now we define $$n_0:=\max\{n_1,n_2\}$$. With this in mind we apply Abel's lemma (summation by parts) and it follows for all $$n>m>n_0$$:

$$|\sum\limits_{k=m+1}^{n} a_kb_k|=|A_nb_n-A_mb_m+\sum\limits_{k=m}^{n-1} A_k(b_k-b_{k+1})|\leq |A_nb_n-A_mb_m|+\sum\limits_{k=m}^{n-1} |A_k(b_k-b_{k+1})| \cdots$$ If $$\left(b_k\right)_{k\in\mathbb{N}}$$ is monotonically decreasing it follows: $$\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_k-b_{k+1})\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_k-b_{k+1})=\frac{\epsilon}{2}+A (b_m-b_n)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.$$

If $$\left(b_k\right)_{k\in\mathbb{N}}$$ is monotonically increasing it follows: $$\cdots<\frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} |A_k|(b_{k+1}-b_k)\leq \frac{\epsilon}{2}+ \sum\limits_{k=m}^{n-1} A(b_{k+1}-b_k)=\frac{\epsilon}{2}+A (b_n-b_m)<\frac{\epsilon}{2}+\frac{\epsilon A}{2A}=\epsilon.$$ So in both cases $$\sum\limits_{k=1}^{\infty} a_kb_k$$ satisfies the Cauchy criterion and hence is convergent.

Is this correct or is there a more elegant/faster approach?

• This is nicely done. (+1) There are other proofs I know but I don't think they are more elegant or faster. – RRL Nov 9 '20 at 0:57

For another approach, we can show that the sequence of partial sums converges without using the Cauchy criterion. We have existence of the limits $$\lim_{n \to \infty}A_n = \lim_{n \to \infty}\sum_{k=1}^na_k =A$$ and $$\lim_{n\to \infty}b_n = b .$$

Summing by parts, we get

$$S_n =\sum_{k=1}^n a_kb_k = a_1b_1+\sum_{k=2}^n (A_k - A_{k-1})b_k = a_1b_1+\sum_{k=2}^{n} A_k b_k- \sum_{k=2}^{n} A_{k-1} b_k \\ = \sum_{k=1}^{n} A_k b_k- \sum_{k=1}^{n-1} A_{k} b_{k+1} = A_nb_{n+1} + \sum_{k=1}^{n} A_k (b_k - b_{k+1})$$

The series $$\sum(b_k - b_{k-1})$$ converges since $$\sum_{k=1}^n (b_k - b_{k+1}) = b_1 - b_{n+1} \to b_1 - b$$ as $$n \to \infty$$. Since $$(A_k)$$ is a bounded sequence and the terms $$(b_k- b_{k+1})$$ are all of the same sign, it follows that $$\sum A_k(b_k - b_{k+1})$$ is convergent.

Therefore, the series $$\sum a_kb_k$$ converges since

$$\sum_{k=1}^\infty a_kb_k = \lim_{n \to \infty}A_nb_{n+1} + \lim_{n \to \infty}\sum_{k=1}^{n} A_k (b_k - b_{k+1}) = Ab + \sum_{k=1}^\infty A_k(b_k - b_{k+1})$$

It follows from Dirichlet's criterion. Indeed, suppose $$b_k \le b_{k+1}$$ and $$\lim b_k = b$$.

Then

\begin{align} \sum_{k=1}^n a_k b_k &= \sum_{k=1}^n a_k b - a_k (b - b_k) \\ &= b\sum_{k=1}^n a_k - \sum_{k=1}^n a_k (b - b_k). \end{align}

The first sum converges as it is a hypothesis and the second sum's convergence follows from the aforementioned theorem.

But this is not an improvement over your argument since the usual proof of Dirichlet's criterion is done through summation by parts.