Show that the product of the series with itself is not absolutely convergent. $$\sum_{1}^{\infty}\frac{(-1)^n}{\sqrt{n}}$$
I tried looking at Cauchy product series.
$$c_n= a_1b_n+...+a_nb_1\implies c_n= (-1)^n[\sum_{1}^{n}\frac{1}{\sqrt{n+1-r}\sqrt{r}}]\implies c_n\geq(-1)^n.n.\frac{1}{\sqrt{n}.\sqrt{n}}$$
the resultant series $\sum(-1)^n$ is an alternating series. I doubt if I can conclude anything from this as the terms of the series are not non negative. Moreover is the original series is not absolutely convergent, I am not sure if looking at Cauchy product is same as looking at the product itself.
Note that$$|c_n|=\frac1{\sqrt n}+\frac1{\sqrt2}\frac1{\sqrt{n-1}}+\cdots+\frac1{\sqrt n}\geqslant\frac n{\sqrt n}=\sqrt n.$$