I was trying to prove Abel's Test for convergence and noticed that
Since $\sum\limits_{k=1}^\infty b_k$ converges ,then for all $\varepsilon>0$ $\exists N_1\in \mathbb{N}$ such that for all $n,m \ge N_1$ $\left|\sum\limits_{k=n}^m b_n\right|< \varepsilon $ , Since $a_n $ converges ,then for all $\varepsilon>0$ $\exists N_2\in \mathbb{N}$ such that for all $n\ge N_2$ $|a_n -a|< \varepsilon$ choose $N= \max\{N_1 , N_2\}$ , then $\forall n,m\ge N$
$$\left|\sum\limits_{k=n}^m b_n\right|(a-\varepsilon)<\left|\sum\limits_{k=n}^m b_n a_n\right|<\left|\sum\limits_{k=n}^m b_n\right|(a+\varepsilon)$$
Best Answer
Your conjecture is not true because modifying any of the $a_k$ does not affect the limit $a$, but changes the value of $\sum a_n b_n$ if the corresponding $b_k$ is not zero.
A simple counterexample is $a_n = 0, 1, 1, 1, \ldots$ and $b_n = 1, 0, 0, 0, \ldots$. Then $a_n \to a = 1$, $\sum_{n=1}^\infty b_n = b = 1$ and $\sum_{n=1}^\infty a_n b_n = 0 \ne ab$.