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