Four very closely related questions on "crossing the border" between convergent/divergent positive real series. Technically there's four questions here, although they are very closely related, and I don't see any value in asking them separately. They're, in some sense, to do with "crossing the border/barrier" of convergence/divergence, which is sort of only a thing in some contexts but not in others.
Suppose $\ a_n>0\ \forall\ n\in\mathbb{N}\ $ and $\ \displaystyle\sum a_n\ $ diverges.
I was wondering if this statement is true:
$\ \displaystyle \sum n^{\alpha} a_n\ $ converges for all $\ \alpha<0,\ \implies\ \displaystyle \sum {a_n}^{\ \beta}\ $ converges for all $\ \beta>1\ ?$
and what about the converse statement?
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Similarly, suppose $\ a_n>0\ \forall\ n\in\mathbb{N}\ $ and $\ \displaystyle\sum a_n\ $ converges.
Then what about this:
$\ \displaystyle \sum n^{\alpha} a_n\ $ diverges for all $\ \alpha>0,\ \implies\ \displaystyle \sum {a_n}^{\ \beta}\ $ diverges for all $\ \beta<1\ ?$
and what about the converse of this statement?
 A: All the statements are true and easy to prove if $a_n$ decreases (to zero).
If $a_{2^m}=1$ and $a_n=1/n^2$ otherwise, clearly the first implication is false since $\sum n^{\alpha} a_n=\sum 2^{m\alpha}+$ convergent so is convergent but $a_n^{\beta}$ doesn't go to zero etc (and of course we can tweak this to get $a_n \to 0$ by say putting $a_{2^m}=\frac{1}{\sqrt m}, m \ge 1$ when the second series still diverges for $\beta <2$)
However, by Holder inequality, we have that if we choose $p>1$ st $p|\alpha|>1$ then with $q>1$ its dual so $1/p+1/q=1$ we have $$\sum_{n=1}^N n^{\alpha} a_n \le (\sum_{n=1}^N n^{p\alpha})^{1/p}(\sum_{n=1}^N a_n^{q})^{1/q} \le \zeta(p|\alpha|)^{1/p}M_q^{1/q}$$ where $\sum_{n=1}^{\infty} a_n^{q}=M_q<\infty$ by hypothesis, so the converse implication in the first statement holds.
Similarly if we choose $a_{2^m}=1/m^2, m \ge 1, a_n=1/{n^2}$ otherwise we have that $\sum n^{\alpha}a_n \ge \sum 2^{m\alpha}/m^2$ so diverges, but $\sum a_m^{\beta}$ converges for $\beta>1/2$ so the first implication in the second statement is not true
But the second implication in the second statement is true since applying Holder with $p=\alpha/2+1, q$ its dual, we get $$(\sum_{n=1}^N n^{\alpha}a_n)^{1/(\alpha/2+1)}(\sum_{n=1}^N n^{-q\alpha/(\alpha/2+1)})^{1/q}\ge \sum_{n=1}^N a_n^{1/(\alpha+1)}$$
But $1/q=1-1/p=\alpha/(\alpha+2)$ so $q\alpha/(\alpha/2+1)=2$ hence $(\sum_{n=1}^N n^{-q\alpha/(\alpha/2+1)})^{1/q}$ is bounded independently of $N$ hence $\sum_{n=1}^N n^{\alpha}a_n$ must be unbounded if $\sum_{n=1}^N a_n^{1/(\alpha/2+1)}$ is per hypothesis with $\beta =1/(\alpha/2+1)$
So in conclusion both reverse implications are true in general, while the direct ones require extra hypotheses like $a_n$ decreasing (to zero)
Edit later: per comment the proofs that both pairs of statements are equivalent when $a_n$ decreases to $0$
Let's assume $a_n$ decreases to $0$, $\sum a_n$ diverges and $\sum n^{\alpha} a_n < \infty$ for all $\alpha<0$; but then since $n^{\alpha} a_n$ for $\alpha<0$ also decreases, we have that $n^{1+\alpha}a_n \to 0$ so $a_n \le C_{\alpha}n^{-1-\alpha}$ or $a_n^{\beta} \le C_{\alpha}^{\beta}n^{(-1-\alpha)\beta}$ but for any $\beta>1$ we can find $\alpha <0$ st $(1+\alpha)\beta >1$ so clearly $\sum a_n^{\beta} << \sum n^{(-1-\alpha)\beta} < \infty$; the converse implication we proved above, but in this case a similar proof uses that now $a_n^{\beta}$ decreases to $0$ and the series converges so $na_n^{\beta} \to 0$ from which we immediately that for $\alpha<0$ we can find $\beta>1$ st $\sum n^{\alpha} a_n$ is majorized by sum $\sum 1/n^{1+\delta}, \delta>0$
(using that $\sum b_n< \infty$ and $b_n$ decreasing to $0$ means $nb_n \to 0$ by noting that $Nb_{2N} \le \sum_{N \le n \le 2N}b_n \to 0$)
Same for the second part, where if say there is $\beta <1$ for which $\sum a_n^{\beta}<\infty$ obviously $\beta>0$ so $a_n^{\beta}$ still decreasing hence $na_n^{\beta} \to 0$ so $na_n^{\beta} \le C$ so $n^{1/\beta}a_n \le C_1$ and hence $n^{\alpha}a_n \le C_1n^{-1/\beta+\alpha}$ and we can find $\alpha>0$ small enough st $-1/\beta+\alpha<-1$ making $\sum n^{\alpha}a_n$ convergent hence contradicting assumption on LHS which shows that LHS implies RHS; the other implication was again shown in general, but can also be done along these lines as above.
