Absolute convergence of a series

Let $y_n , \; n \in \mathbb{N}$ be a sequence of real numbers such that for all real valued sequences $x_n , \; n \in \mathbb{N}$ such that $\lim x_n =0$ the series $\sum \limits_{n=1}^{\infty} x_n y_n $ converges. Does it necessarily follow that the series $\sum \limits_{n=1}^{\infty} \left| y_n \right|$ converges?

Solution

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.

The above solution was given by Demetres at the greek forum mathematica.gr