Show that $\sum\frac{(-1)^{n+1}} {{n}^r} \sum\frac{(-1)^{n+1}} {{n}^s} $ by Abel's rule forms a series that doesn't converge when r+s=1. It is a similar problem to that in 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.. It may provide some hint to the latter.
Show that $\frac{1} {{1}^r} -\frac{1} {{2}^r} +\frac{1} {{3}^r} +\dots$ and $\frac{1} {{1}^s} -\frac{1} {{2}^s} +\frac{1} {{3}^s} +\dots$, where 0 < r < 1, being multiplied by Abel's rule, forms a series (say $\sum \nu_n$) that doesn't converge, when r+s=1.
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.
Initial steps are similar to those in the post,
$(\frac{1} {{1}^r} -\frac{1} {{2}^r} +\frac{1} {{3}^r} +\dots)(\frac{1} {{1}^s} -\frac{1} {{2}^s} +\frac{1} {{3}^s} +\dots)\\
=\frac{1} {{1}^r}\frac{1} {{1}^s}+\dots
+[(-\frac{1} {{1}^r} \frac{1} {{(2k)}^s}+\frac{1} {{1}^r} \frac{1} {{(2k+1)}^s}
-\frac{1} {{2}^r} \frac{1} {{(2k-1)}^s}+\frac{1} {{2}^r} \frac{1} {{(2k)}^s}+\dots
-\frac{1} {{k}^r}\frac{1} {{(k+1)}^s}+\frac{1} {{k}^r}\frac{1} {{(k+2)}^s}
-\frac{1} {{(k+1)}^r}\frac{1} {{k}^s}+\frac{1} {{(k+2)}^r}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^r}\frac{1} {{1}^s}+ \frac{1} {{(2k+1)}^r}\frac{1} {{1}^s})
 +\frac{1} {{(k+1)}^{r+s}}]+\dots,$
where
$\sum_{m=1}^{2k}|(-\frac{1} {m^r} \frac{1} {(2k+1-m)^s}+\frac{1} {m^r} \frac{1} {(2k+2-m)^s})|
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{1} {(2k+1-m)^s}(1-\frac{1} {(1+\frac{1}{2k+1-m})^s})\\
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{1} {(2k+1-m)^s}(s\frac{1}{2k+1-m}+O(\frac{1}{(2k+1-m)^2}))
=\sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}},$
for $1-(1+x)^{-s}=-\frac{(-s)}{1!}x-\frac{(-s)(-s-1)}{2!}x^2+\dots.$ We can not easily use $\frac{1}{\sqrt{ab}}>\frac{1}{a+b}$ here, instead we use Taylor expansion. It seems the above sum approximates $\sum_{m=1}^{2k}\frac{1} {k^r} \frac{1} {(k)^{s+1}}\approx \frac{k}{k^{r+s+1}}=\frac{1}{k},$ and so the series (say $\sum \psi_n$) it forms diverges as well.
But here we are gonna show that the serives diverges more than $\sum \frac{1}{k+1}$, which we can't yet show in the above post.
Let s go near 1-0 (i.e. r+s-0), then $\sum \psi_n$ goes near
$\sum_{m=1}^{2k}\frac{1} {m^0} \frac{r+s} {(2k+1-m)^{r+s+1}}
=\sum_{m=1}^{2k}\frac{1} {(2k+1-m)^{2}}=\frac{1}{(2k)^2}+\frac{1}{(2k-1)^2}+\dots+\frac{1}{1^2}>\frac{2}{k+1}$ (A note for myself: for calculation of this sum and the sum in the above post, see Formula for $\frac{1}{(n)^2}+\frac{1}{(n-1)^2}+\dots+\frac{1}{1^2}$.. According to results there, the left side tends to $\frac{\pi^2}{6}$ which is obviously larger than the right side that tends to 0. So $\sum \nu_n\approx \sum_{k=0}^\infty \frac{\pi^2}{6}$, it doesn't oscillates between two values but oscillates to infinity.)
when k $\geq$ 3 (i.e. $\frac{2}{k+1}\leq \frac{1}{2^2}$). Therefore when s is near 0, $|\sum \nu_n|>\sum\frac{2}{k+1}-\sum\frac{1}{k+1}$, which diverges.
My question is how to, in general, prove that it the series $\sum \nu_n$ diverges?
 A: So here we have, for example,
if r<s,
$
|-\frac{1} {{1}^r} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^r} \frac{1} {{(2k-1)}^s}+\dots
-\frac{1} {{k}^r}\frac{1} {{(k+1)}^s}
-\frac{1} {{(k+1)}^r}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^r}\frac{1} {{1}^s}|>
|-\frac{1} {{1}^s} \frac{1} {{(2k)}^s}
-\frac{1} {{2}^s} \frac{1} {{(2k-1)}^s}+\dots
-\frac{1} {{k}^s}\frac{1} {{(k+1)}^s}
-\frac{1} {{(k+1)}^s}\frac{1} {{k}^s}
\dots-\frac{1} {{(2k)}^s}\frac{1} {{1}^s}|>
\sum (\frac{2}{2k+1})^{2s}>\sum (\frac{2}{2k+1})^1
,$
which doesn't coverge to 0. Similar for 2k+1.

For a general solution:
According to https://math.stackexchange.com/a/3787267/577710, (well, it seems to require some modifications, and so does what follows.)
$(n-1)^{1-\max(r,s)} \leq \sum_{m=1}^{n-1}\frac{1}{m^r}\frac{1}{(n-m)^s}
$, and so $s<(2k)^{1-s}s<s\cdot(2k+1-1)^{1-\max(r,s)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$, and so the series (at index n being odd) approximates $\sum_{k=0}^\infty\frac{1}{k+1}-\sum_{k=0}^\infty s$ which diverges to infinity, which fits our discussion for s=1.
When r=s=1/2, we see $\sqrt{2k}s<s\cdot(2k+1-1)^{1-\max(r,s)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$. And so the series diverges (oscillates actually) even more, not (as I thought in https://math.stackexchange.com/a/3787073/577710) oscillating between two values.

Correction:
$(n-1)(\frac{2}{n})^{2\max(r,s)} \leq \sum_{m=1}^{n-1}\frac{1}{m^r}\frac{1}{(n-m)^s}
$, so
$\frac{2s}{{(k)}^{2s+2}}\approx s\cdot (2k)(k+1/2)^{-2\max(r,s+1)} \leq \sum_{m=1}^{2k}\frac{1} {m^r} \frac{s} {(2k+1-m)^{s+1}}$.
(In this calculation we omit the fact that r and s+1 are not symmetrical; as a result we have not 1+2...+m+...2k which equals to O(k^2) and so the lower bound is very small.)
Therefore it's clear (but not from the above inequality) that when s is near 0, the series diverges as described above, but it is still unclear that when s=r=1/2, the series diverges to infinity or oscillates between two values.
