A problem of absolute convergence

[r9m]

Let $\displaystyle \{a_n\}_{ n\ge 1}$ be a real valued sequence, such that for every real convergent sequence $\{r_n\}$, the series $\displaystyle \sum\limits_{n=1}^{\infty} r_na_n$ converges. Does it follow that $\displaystyle \sum\limits_{n=1}^{\infty} |a_n| < \infty$ (converges absolutely)?

[Tolaso J Kos]

A solution Demetres gave at mathematica.gr

Suppose on the contrary that the series $\sum \limits_{n=1}^{\infty} \left|y_n\right|$ diverges. Define the sequence of the natural numbers $n_0, n_1, n_2, \dots$ as follows: Set $n_0=1$. Having defined $n_0, n_1, n_2 , \dots$ we define $n_{k+1}$ to be the least natural $m$ that is greater than $n_k$ and satisfies

$$\sum_{r=n_k+1}^{m} \left|y_r\right|>1$$

The sequence is well defined since the series diverges. For $n_k < i \leq n_{k+1}$ we define $\displaystyle x_i =\frac{{\rm sign}(y_i)}{k+1}$. Then $x_n \rightarrow 0$. Furthermore,

$$\sum_{i=n_k+1}^{n_{k+1}} x_iy_i = \sum_{i=n_k+1}^{n_k} \frac{|y_i|}{k+1} > \frac{1}{k+1}$$

implying that the series $\sum \limits_{n=1}^{\infty} x_n y_n$ diverges. A contradiction. So, we conclude that the series $\sum \limits_{n=1}^{\infty} \left| y_n \right|$ converges.

Note: Here also $a_n$ is replaced by $y_n$ and $r_n$ by $x_n$.

[r9m]

Cool solution T!