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An absolutely convergent series

Let \{y_n\}_{n \in \mathbb{N}} be a sequence of real numbers such that for all real sequences \{x_n\}_{n \in \mathbb{N}} with \lim \limits_{n \rightarrow +\infty} x_n=0 the series \sum \limits_{n=1}^{\infty} x_n y_n converges. Prove that the series \sum \limits_{n=1}^{\infty} |y_n| also converges.

Solution

Suppose on the contrary that the series diverges. Define the sequence of the naturals n_0,n_1,n_2,\ldots as follows. We set n_0=1. Having defined n_0, n_1, \dots, n_k we define n_{k+1} to be the least natural m that is greater than n_k and satisfies \sum \limits_{r=n_k+1}^{m} |y_{r}| > 1. The sequence is well defined because the series \sum \limits_{n=1}^{\infty} |y_n| diverges.

For n_k < i \leq n_{k+1} we define \displaystyle x_i = \frac{\mathrm{sign}(y_i)}{k+1}. Then we note that x_n \rightarrow 0 and

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

thus the series \sum \limits_{n=1}^{\infty} x_n y_n diverges. This leads to contradiction hence the result.

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