I am wondering if the following is true :
Let $(u_n)_{n \in \mathbb{N}}$ be a decreasing sequence such that $u_n \rightarrow 0$ as $n \to \infty$. Let $(a_n)_{n \in \mathbb{N}}$ such that $(a_n)_n \in [-A,A]^{\mathbb{N}}$ is dense in $[-A, A]$ for some $A \in \mathbb{R}_{>0}$ then: $$\sum_{n=0}^\infty u_na_n$$ converge.
I am enable to find a counter exemple. But I've notice the following :
- If the sequence : $\sum_{k=0}^N a_k$ converge then it's bounded and maybe we can use an Abel transform to prove that it converges.
- When we take $a_n = \sin(n)$ (then $A = 1$) then it converges. We can use an Abel transform to prove it.
Possible answer too long for the comments:
Is this a counter example: take $c_n$ to be the positive rationals ordered $$\frac{1}{1}, \frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3}, \cdots .$$ The rational $n/1$ then arises at the position $n(n-1)/2 + 1$. Now create $a_n$ by inserting each negative rational, taken in the same order, immediately after $c_{2^k}$.
This will increase the position index of $n/1$ in $\{a_n\}$ from $n(n-1)/2 + 1$ to something like $n(n-1)/2 + \log_2 (n(n-1)/2 + 1)$. Now take $u_k$ as slowly decreasing, for instance $1/(\sqrt[3]{k})$.
I think then as you sum $u_k a_k$ you will have infinitely growing contributions from the positive rationals $n/1$, each divided by $\sqrt[3]{ n(n-1)/2 + \log_2 (n (n-1)/2 + 1)}$. The sum will be offset by negative rationals, but arising so infrequently that the sum must still diverge. Apologies for not formalising this; please say if you think this it works.
If the requirement that $a_n$ be bounded I believe a similar approach can be taken. Start with a complete sequence of the positive rational in the interval $[0,A]$. Then add the negative ratioals, using the same sequence but include each positioned as the $1, 2, 2^2, 2^3, \cdots $ term. Finally mix in the constant $A/2$ as every other element. Thus if $r_n$ is a list of the positive rationals, $$a_n = \left\{ \begin{array}{ll}A/2 & \text{when } n= 2,4,6, \cdots \\ -r_k & \text{when } n = 1, 3, 5, 9, 17, 33, \cdots, 2^k+1, \cdots \\ r_k &\text{for other } n, n = 7, 11, 13, 15, 19, 21, 23 ,\cdots \end{array} \right.$$ and choose $p_n$ to be any slowly declining positive sequence with unbounded sum.
Define $$u_n = \left\{ \array{ p_{n/2} &\text{for even } n \\ u_{n+1} & \text{for odd n}}\right. $$ so that $\{u_n\} = p_1, p_1, p_2, p_2, \cdots $. I believe then $\sum_n u_na_n $ is also unbounded.