I am trying to show that the series $\dfrac {1} {\sqrt {1}}-\dfrac {1} {\sqrt {2}}+\dfrac {1} {\sqrt {3}}-\ldots $ is convergent, but that its square (formed by Abel's rule) $\dfrac {1} {1}-\dfrac {2} {\sqrt {2}}+\left( \dfrac {2} {\sqrt {3}}+\dfrac {1} {2}\right) -\left( \dfrac {2} {\sqrt {4}}+\dfrac {2} {\sqrt {6}}\right) -\ldots $ is divergent.
Now in order to verify the author's claims. I observed that the first series $\sum _{n=1}^{n=\infty }\dfrac {\left( -1\right) ^{n+1}} {\sqrt {n}}$ converges by Leibniz test or alternating series test, but i am having a hard time firstly verifying the product would create the second series and then finding a generic pattern in it to be able to test for convergence.
Any help would be much appreciated.
Edit I think the product by abels' rule should yield. $\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{n+1}} {\sqrt {n}}.\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{n+1}} {\sqrt {n}} = \lim _{p\rightarrow \infty }\sum _{n=1}^{n=\infty } \frac {\left( -1\right) ^{n+1}} {\sqrt {n}}.\frac {\left( -1\right) ^{p-(n+1)}} {\sqrt {p - n}}= \lim _{p\rightarrow \infty }\sum _{n=1}^{n=\infty }\frac {\left( -1\right) ^{p}} {\sqrt {p - n}\sqrt {n}}$