I found this relationship in a probability book (Durrett, "Probability: Theory and examples", 4th edition; Theorem 2.5.8), though it is just simple calculus:
$$\sum_{m=n}^\infty m^{-2/p} \leq \frac{p}{2-p}(n-1)^{(p-2)/p} \leq Cy^{p-2},$$
where $y \in [(n-1)^{1/p}, n^{1/p}]$. Here, $p \in (1,2)$.
The first inequality comes from comparing the sum with an integral ($\sum_{m=n}^\infty m^{-2/p} \leq \int_{n-1}^\infty x^{-2/p}$), I understand that part. But if $y \in [(n-1)^{1/p}, n^{1/p}]$ and being $f(x) = x^{p-2}$ decreasing, isn't it the case that $\frac{p}{2-p}(n-1)^{(p-2)/p} \geq Cy^{p-2}$? I am assumig that $C$ does not depend on $n$.
I am sorry for asking such an elementary question, but I have spent a bit of a time stuck in that second inequality and cannot find what is wrong with my reasoning.
Your question: Does there exists $C > 0$ such that $$\forall\,y \in [(n-1)^{1/p}, n^{1/p}], \frac{p}{2-p}(n-1)^{(p-2)/p} \leq Cy^{p-2}?$$
To make things easier, let's move $y^{p-2}$ to LHS, so that every quantity in the inequality is in the form of $\pm\underbrace{(\dots)}_{\text{positive}}$.
$$\frac{p}{2-p} \frac{(n-1)^{-(2-p)/p}}{y^{-(2-p)}} = \frac{p}{2-p} \frac{y^{2-p}}{(n-1)^{(2-p)/p}} \le \frac{p}{2-p} \underbrace{\frac{n^{(2-p)/p}}{(n-1)^{(2-p)/p}}}_{\to 1}$$
Since convergent sequence is bounded, simply take $C$ to be any number larger than or equal to $\sup_n (n/(n-1))^{(2-p)/p}$.