Let $a_1,\dots,a_n,\dots$ be a sequence of positive reals such that $a_i<a_j$ if $i<j$ and let $ 0<\beta<2 $. Suppose there exists $0<m<M$ such that $m \leq a_{i+1}-a_i \leq M$ for every $i$.
Let $\delta>0$ and $\alpha>0$. For what values of $n\in \mathbb{N}$ and $m,M$ do we have that $$ \frac{\sum_{i=1}^n (a_{i+1}^{\alpha}-a_i^{\alpha})^{\beta}}{ \left(\sum_{i=1}^n a_{i+1}^{\alpha}-a_i^{\alpha}\right)^{\beta} } \leq \delta? $$
Obviously, for all $n\geq 2 \in \mathbb N$ ,
$\delta=(n-1)$, for $\alpha>0$.
Say, $(a_{i+1}-a_{i})=d_i$, and we know $0<m\leq d_i \leq M$.
So, the given sum can be rewritten as $\frac{\sum_{i=1}^{n-1} {d_i}^{\alpha}}{(\sum_{i=1}^{n-1} {d_i})^{\alpha}}$
W.L.O.G let $d_1$ is the biggest term among these. Then we can write the expression as
$\frac{1+{x_1}^\alpha + {x_2}^\alpha + {x_3}^\alpha+....{x_{n-2}}^\alpha}{(1+{x_1}+{x_2}+....+{x_{n-2}})^\alpha}$
Where, $x_i=\frac{d_{i+1}}{d_1}$ and $x_i<1$, so $\sum x_i \leq n$. Now we don't need any data about $m, M$.
From here we can see if $\alpha>1$ the whole expression becomes less than 1 and goes towards 0 with increase in $\alpha$ unless all $x_i$ aren't 1. In the later case the value of the expression is $n-1$.
Now, if $\alpha\leq1$ and goes to zero, the value of the expression increases and converges to $n-1$ for all values of $x_i$. So, $\delta=n-1$.