<s>[INST] Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn't. Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ (say $\sum u_n$) converges, and its square  (say $\sum v_n$)(formed by Abel's rule) doesn't.
Abel's rule: given $\sum a_n, \sum b_n$, $\sum_{n=0} ^\infty c_n=\sum_{n=0} ^\infty [\sum_{i=0} ^n a_{n-i}b_i]$ is the infinite series gotten from multiplication of two series.
According to this rule,
$(\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots)^2 
=\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{1}}]+\dots
-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}]
+[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}}]\dots,$
which, if we sum by adding nearby items first, equals
$\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[(-\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}
-\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
-\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}
-\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}
\dots-\frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}})
+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]<\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[
\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]=\sum_{k=0}^\infty \frac{1}{k+1}$ which diverges.
But in order to prove the series diverges we probably need to prove it's no less than another divergent series.
It's possibly not complicated. I will see how I can modify the proof to make it work. Perhaps I need to use the ratio test $\frac{c_{n+1}}{c_n}=1+\frac{A}{n}+O(\frac{1}{n^2})$, considering the series approximates $\sum_{k=0}^\infty \frac{1}{k+1}$
Context: the significance of the statement is that if it is true, then $\lim_{x\to 1}v_n x^n \neq c_n$, and so (though, for $\sum u_n x^n$ converges absolutely, we have $(\sum u_n x^n)^2=\sum v_n x^n$, letting $x\to 1$,) we don't have $(\sum u_n)^2$ (i.e. the limit of the left side) equals $\sum v_n$.
 [/INST] I'd call that the Cauchy product.
To show that your series diverges, it suffices to use what might be the bluntest tool available, the $n$th term test.
The $n$th term of the "product" is given by
$$\pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{n-i}}\frac{1}{\sqrt{i}} = \pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}.$$
We want to show that the series diverges by showing that these don't converge to $0$, which we can do by finding a lower bound on the absolute value.The terms in the sum are minimized when $i(n-i)$ is maximized, and we know that the function $f(x) = x(n-x)$ is maximized at $x=n/2$ (it is a downard pointing parabola so the maximum is at the critical point, which can be identified with the derivative).
All the terms in the sum are nonnegative, so the absolute value won't cause trouble,
$$\left|\pm \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}\right| = \left|\sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}\right| = \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}},$$
and the bound is then easy to apply,
$$\geq \sum_{i=1}^{n-1} \frac 1{\sqrt{(n/2)^2}} = \sum_{i=1}^{n-1} \frac 2 n = \frac{2(n-1)}{n}.$$
The latter is greater than or equal to $1$ for all $n\geq 2$.
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