Seeking for a hint to a limit question $\lim_{n \to \infty}(a_{n+1}^{\alpha}-a_n^{\alpha})=0$ Assume ${a_n}$ is a positive, strictly increasing sequence, while $a_{n+1}-a_n$ is bounded.
Prove: for every real number $\alpha \in (0,1)$, $$\lim_{n \to \infty}(a_{n+1}^{\alpha}-a_n^{\alpha})=0.$$
I'm wondering how can I associate $a_{n+1}-a_n$ with $a_{n+1}^{\alpha}-a_n^{\alpha}$.
 A: As indicated in the comments, there are two cases to consider.


*

*$a_n$ is bounded. Since it is increasing it converges to some $a>0$ and then $\lim_{n\to\infty}(a_{n+1}^\alpha-a_n^\alpha)=a^\alpha-a^\alpha=0$. 

*$a_n$ is anbounded, and $\lim_{n\to\infty}a_n=+\infty$. Use the inequlity
$$
x^\alpha-y^\alpha\le\frac{x-y}{x^{1-\alpha}},\quad 0\le y\le x,
$$
which you can prove without using the MVT.

A: We are going to deke around the MVT in the case where $(a_n)_{n \in N}$ does not converge.......(1)For  positive integers $ m,n$ with $m<n$ we have $$m^{-j} \binom  {m} {j} \le n^{-j} \binom {n} {j} $$ for $ 0\le j$, from the def'n of the binomial co-efficient, so $$\binom {m}{j}\le (m/n)^j \binom {n}{j}. $$.From this,  , by  comparing term-by-term the binomial expansions of $(1+z)^m$ and $(1+(m/n)z)^n$ we deduce that when $z \ge 0$ we have $$(1+z)^m \le (1+(m/n)z)^n$$ so$$(1+z)^{m/n} \le 1+(m/n)z.$$  By continuity  we have $$(1+z)^{\alpha}\le 1+ \alpha z$$ whenever $z \ge 0$ and $0<\alpha <1$.(This inequality has many uses.)......(2) Suppose your sequence $(a_n)_{n \in N}$ increases to $\infty.$ For some $ k>0$, we have $k>a_{n+1}-a_n>0$ for all $n$...... Now let $$ z_n =a_{n+1}/a_n-1.$$We have $$k/a_n>z_n>0.$$Now we obtain  $$0<a_{n+1}^{\alpha}-a_n^{\alpha}$$ $$=a_n^{\alpha} ((1+z_n)^{\alpha}-1)$$ $$\le a_n^{\alpha}(1+\alpha z_n)-1)$$ $$=a_n^{\alpha} . \alpha. z_n$$ $$<a_n^{\alpha}.\alpha.(k/a_n)$$ $$=\alpha k /(a_n)^{1-\alpha}$$ which goes to $0$ as $n \to \infty$. The case where $(a_n)_{n \in N}$ has a limit has been covered in another answer.
