Is it true that if $a_n \to a$ and $\sum b_n \to b$ then $\sum a_n b_n \to ab$

Question

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 $\forall n,m\ge N$

$$\left|\sum\limits_{k=n}^m b_n\right|(a-\varepsilon)<\left|\sum\limits_{k=n}^m b_n a_n\right|<\left|\sum\limits_{k=n}^m b_n\right|(a+\varepsilon)$$

Answer 1

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 $\sum_{n=1}^\infty a_n b_n = 0 \ne ab$.

Answer 2

It is certainly not true, as commented by Martin R. Let's take an (non-trivial) example: $a_n = 1/n$ and $b_n = 1/n^2$, then it is well-known that $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} = \zeta(2)$$ and $a_n \to 0$, but $$\sum_{n=1}^{\infty}a_n b_n = \sum_{n=1}^{\infty}\frac{1}{n^3} = \zeta(3) \sim 1.202$$ (no need to compute $\zeta(3)$, it is enough to know that the series is convergent and non-zero).

Answer 3

An emblematic counter-example can be made with the infinite series $$ \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots = \frac{1}{1}+\frac{1}{1}+\frac{1}{2\cdot 1}+\frac{1}{3\cdot 2 \cdot 1}+\frac{1}{4\cdot 3\cdot 2 \cdot 1}+\ldots $$ whose limit is the Euler number $e=2.71828\ldots $ If $b_{n}=\frac{1}{n!}$ and $a_{n}=\frac{1}{n+1}$ we have $$ \sum_{n=1}^{\infty}b_{n} = b_{1}+b_{2}+\ldots+b_{n}+\ldots = \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots +\frac{1}{n!}+\ldots = e-1 $$ and $$ \sum_{n=1}^{\infty}a_{n}\cdot b_{n} = a_{1}\cdot b_{1}+a_{2}\cdot b_{2}+\ldots+a_{n}\cdot b_{n}+\ldots \\ = \frac{1}{2\cdot 1!}+\frac{1}{2\cdot2!}+\frac{1}{4\cdot3!}+\ldots +\frac{1}{(n+1)\cdot n!}+\ldots = e-1-1 $$ But $\frac{1}{n+1}\rightarrow 0$, $\left(\sum_{n=1}^{\infty}\frac{1}{n!}\right)\rightarrow e-1$ and $0\cdot ( e -1) \neq (e -1 -1)$.