# Boundedness of a real sequence.

Let $\{a_n\}$ be a sequence in $\mathbb{R}$ such that $\sum |a_n||x_n| < \infty$ whenever $\sum |x_n| < \infty$. Prove that $\{a_n\}$ is bounded.

My Attempt :

We have to show $\exists$ $B \in \mathbb{R}^+$ s.t. $|a_n| < B$.

$\sum |x_n| < \infty$ gives $\lim |x_n| = 0$ i.e. for any $\epsilon > 0$ $\exists$ $m_1 \in \mathbb{N}$ s.t. $|x_n| < \epsilon$ when $n > m_1$

$\sum |a_n||x_n| < \infty$ gives $\lim |a_n x_n| = 0$ i.e. for any $\epsilon > 0$ $\exists$ $m_2 \in \mathbb{N}$ s.t. $|a_nx_n| < \epsilon$ when $n > m_2$.

I can not approach any more. I require $|x_n| > C$ $\forall n > m$ for some real $C$ and natural number $m$ to prove $a_n < B$.

One more question I have in this context. What will be if we do not consider absolute convergence of the series $\sum x_n$ and $\sum a_n x_n$ ?

It might be easiest to approach this by contradiction : Suppose $(a_n)$ is unbounded, then for each $k \in \mathbb{N}, \exists a_{n_k}$ such that $$|a_{n_k}| > 2^k$$ So consider the sequence $(x_n)$ such that $$x_{n_k} = \frac{1}{2^k}, \text{ and } x_n = 0 \text{ otherwise}$$ Then $$\sum|x_n| < \infty, \quad\text{ but }\quad \sum |a_n||x_n| = +\infty$$ This is a contradiction, so $(a_n)$ must be bounded.
• Not that I am aware of. If you knew that $$\sum |a_nx_n| \leq M\sum |x_n|$$ for some $M > 0$, then a direct approach would be possible. – Prahlad Vaidyanathan Nov 25 '13 at 6:35
Well for sure we know that if we don't consider absolute convergence even though $(a_n)_{n\in \mathbb{N}}$ is bounded we can't conclude from $\sum x_n$ converges that $\sum a_n \cdot x_n$ converges.
But that should be the only problem. Because of Leibniz we know that $$\sum (-1)^n \cdot x_n$$ always converge if $x_n$ is monotone decreasing sequence of positive real numbers we can chose $x_n$ to decrease arbitrary slow. But as $$\sum a_n \cdot x_n$$ converges we know that $a_n\cdot x_n$ must converge to zero. When $a_n$ isn't bounded we can find a subsequence which is monotone and not bounded. Now we set $x_{n_k}=\frac{1}{a_{n_k}}$ for this subsequence and the other $x_n$ we chose that we can use Leibniz. But then $a_n\cdot x_n$ has a subsequence which converges to $1$, this is a contradiction.