# Sequence convergence and limit

Determine if the sequence $\left\{(n+1)^{\alpha}-n^{\alpha}\right\}_{n=1}^{+\infty}, 0 < \alpha < 1$ converges or diverges. If it converges, give its limit value.

\begin{align} 0, \ converges \end{align}

I know that this sequence limit can be evaluated easily with some algebra manipulation getting the corresponding function and applying L'Hopital's rule to it. But I would like to try another way around just for the sake of playing with something less mechanical.

Below is my reasoning. I'd be grateful if someone can point me out if it is correct.

I have tried the following:

First, I've evaluated a few terms looking forward to eventually fathom some pattern. So:

$a_{1} = 2^{\alpha}-1 \to 2^{\alpha} = a_{1}+1$

$a_{2} = 3^{\alpha}-2^{\alpha} = 3^{\alpha} - (a_{1} + 1) \to 3^{\alpha} = a_{2} + a_{1}+1$

$a_{3} = 4^{\alpha}-3^{\alpha} = 4^{\alpha} - (a_{2} + a_{1} + 1) \to 4^{\alpha} = a_{3} + a_{2} + a_{1}+1$

$a_{4} = 5^{\alpha}-4^{\alpha} = 5^{\alpha} - (a_{3} + a_{2} + a_{1} + 1) \to 5^{\alpha} = a_{4} + a_{3} + a_{2} + a_{1}+1$

$\vdots$

$a_{n} = (n+1)^{\alpha} - n^{\alpha} = (n+1)^{\alpha} - (a_{n-1} + a_{n-2} + a_{n-3} + \ldots\ + a_{3} + a_{2} + a_{1} + 1)$

As you can see, each new term equals the next index to the alpha-th power subtracted from the sum of all $n-1$ terms plus one.

Subtracting $(n+1)^{\alpha}$ from both sides yields:

$-n^{\alpha} = -(a_{n-1} + \ldots + a_{3} + a_{2} + a_{1} + 1)$

Let $m = a_{n-1} + \ldots + a_{3} + a_{2} + a_{1}, m \in ℝ$ be the sum of $a_{n-1}$ terms. We have:

$n^{\alpha} = m + 1$

If we take the roots on both sides we get:

$n = {(m+1)}^{\frac{1}{\alpha}}$

It is given by the problem that $0 < \alpha < 1$, therefore $\frac{1}{\alpha} > 1$.

Substitute $n$ in our $a_{n}$:

$a_{n} = (n+1)^{\alpha} - n^{\alpha} =$

$= ((m+1)^{\frac{1}{\alpha}}+1)^{\alpha}-((m+1)^{\frac{1}{\alpha}})^{\alpha}$

Since we are dealing with a limit process, we're interested when $m$ is a very large number, that is, when $m \gg 1$. In this particular, the second $+1$ in the first term above renders no effect. Therefore we can discard him. We will get:

$((m+1)^{\frac{1}{\alpha}})^{\alpha}-((m+1)^{\frac{1}{\alpha}})^{\alpha} = 0, m \gg 1$. We can conclude that the limit is $0$, that is:

$\therefore \boxed{\lim_{n\to\infty} (n+1)^{\alpha} - n^{\alpha} = 0}$ and the given sequence converges. $\blacksquare$

This is a comment. For $x>0$ and $0<a<1$ let $f_a(x)=x^a.$ For $n\in \Bbb N$ there exists $y\in (n,n+1)$ such that $$f_a(n+1)-f_a(n)=f'_a(y)=ay^{a-1}.$$ Now $0<ay^{a-1}<an^{a-1}$ and $an^{a-1}\to 0$ as $n\to \infty.$
So $f_a(n+1)-f_a(n)\to 0.$
BTW..... $y>a(n+1)^{a-1}>a/(n+1)$ so $\sum_{n\in \Bbb N}((n+1)^a-n^a)=\infty.$