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 $\sum v_n$.
 A: 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 $n\geq 2$.
A: This is inspired by an answer.
$|-\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}
-\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
-\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}
-\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}
\dots-\frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}}|
\\>
2\sum_{m=1}^k \frac{m}{{\sqrt{m(2k+2-m)}}^3}>2\sum_{m=1}^k \frac{m}{(k+1)^3}= \frac{k(k+1)}{(k+1)^3}=\frac{1}{(k+1)}-\frac{1}{(k+1)^2},$
So $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$ diverges. And so $\sum v_n$ is the sum of two divergent series. Then how to prove that it diverges?
By far we see
$v_n < \frac{1}{k+1}-\frac{1}{(k+1)^2}+\frac{1}{1+k}=-\frac{1}{(k+1)^2}$, and so |$\sum v_n$| is larger than a convergent series, which implies we have not proved the divergence of $\sum v_n$. It seems the problem is more tricky than it looks, were my calculation correct. Perhaps I need to shrink the series $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$less.
I notice when I use $\frac{1}{\sqrt{ab}}>\frac{2}{a+b}$, the difference between the two sides can be large when a, b is very different, e.g. $\frac{1}{\sqrt{1(2k+1)}}$ differs from $\frac{2}{2k+2}$ as much as $O(\frac{1}{\sqrt{k}})$ differs from $O(\frac{1}{k})$, e.g. k=1/1000,000. But does that matter for knowing the divergence of the series?
Again, at least now I realize it's addition of two divergent series instead of a divergent series and a convergent one, and that put me in the right track.

Addition:
Now I realize it's addition of two divergent series instead of a divergent series and a convergent one. My question is then how can one prove that the series is divergent.
Details of my thoughts about this new question is described in my answer.
The problem is trickier than it seems, and possibly it should be, since we are dealing with multiplication of two convergent (though not absolutely) series, and the most natural result is that we get a convergent series, or if in calculation we separate it into several divergent series, most naturally their sum should be a convergent series like $\sum\frac{1}{n^2}$ in my answer.
Thinking of the context I guess the problem has something to do with the difference, or error $\delta$, between $\sum v_nx^n$ and $\sum v_n$ (possibly, when it is divergent, the former doesn't converge uniformly as $n\to \infty$, and at x=1, it doesn't converges to the latter as $x\to 1$). Also it is possibly caused by the way we do multiplication, where the series tail engages in many more multiplications (with each other of tail terms) than the series head. This could amplify the small 'error' of the tail.
Thus a plausible approach is to calculate $(\sum u_nx^n)^2$ first, and then compare it with $\sum v_n$ and see where the error $\delta$ is mostly ignored (in the proof) (and amplified by multiplication) that leads to incompleteness of the proof. Perhaps this way will be more efficient than trying directly to shrink the series $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$ less.
Again, it is important to realize what I get above is two divergent series.

Correction:
I see, (https://math.stackexchange.com/a/3787287/577710)

*

*it's $v_n$ doesn't converges to 0 and so $\sum v_n$ diverges.

*And adding adjacent items follows not the rule.

The series possibly diverges between two numbers. It is loosely 'convergent', considering series 'pausing' at n odd and n even, which fits our intuition that convergent series's product is (though not always strictly) 'convergent'.
This example shows not noticing a simplest fact when stuck and a slightly different understanding of the problem from what it is can delay progress.
