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

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 $$\forall n,m\ge N$$

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

• That cannot be true because modifying finitely many of the $a_n$ does not change the limit $a$, but can change the value of $\sum a_n b_n$. Commented Dec 20, 2023 at 13:56
• FWIW, the error (in the title question) is passing from asymptotic behavior of the tails (the estimates in the post body) to a conclusion about the sum. Commented Dec 20, 2023 at 14:15

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 $$\sum_{n=1}^\infty a_n b_n = 0 \ne ab$$.
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).
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 $$0\cdot ( e -1) \neq (e -1 -1)$$.