Let $\{x_n\}$ be a given series such that it satisfies the following conditions for all sequence $\{y_n\}$ in real numbers converging to $0$. It is given that the sequence $\{y_n\}$ converges to $0$ and the series $\sum_{n=0}^{\infty} x_ny_n$ is convergent. Then show that the series $\sum_{n=0}^{\infty} |x_n|$ is convergent$,$ that is the series $\sum_{n=0}^{\infty} x_n$ is absolutely convergent.
My attempt: Since $\{y_n\}$ is converges to $0$ then for each $ε>0$ there exist a natural number $k_1$ such that $|y_n|< ε$ for all $n>k_1$. Since the series $\sum_{n=0}^{\infty} x_ny_n$ is convergent then the tail of the series goes to zero that is $x_ny_n$ tends to $0$ as $n$ tends to $\infty$. But can not proceed further to complete the proof. Please help me to solve this. Thanks in advance.
Suppose that $\sum |x_n|=\infty$. We will construct a sequence $\{y_n\}$ such that $y_n\to 0$ and $\sum x_ny_n$ diverges.
Since $\sum |x_n|=\infty$ we can find a strictly increasing sequence of natural numbers $N_i$ such that the partial sums satisfy $$\sum_{n=1}^{N_k}|x_n|>k$$
Now define $y_n$ by $$N_k≤n<N_{k+1}\implies y_n=\text {sign} (x_n)\times \frac 1{\sqrt{k}}$$
Clearly $y_n\to 0$. We remark that the partial sum $$\sum_{n=1}^{N_k}x_ny_n≥ \frac 1{\sqrt{k}}\sum_{n=1}^{N_k} |x_n|≥\sqrt { k}$$
Hence the partial sums for $\{x_ny_n\}$ tend to $\infty$ and we are done.