Convergence of $ ( n^{\alpha}-(n+1)^{\alpha} )_{n=1}^{\infty} $

Question

Examine the convergence of the sequence $( n^{\alpha}-(n+1)^{\alpha})_{n=1}^{\infty}$ where $\alpha$ is any number between $0$ and $1$

I was asked the question in an interview. I was already at a loss. Intuitively I thought that for large $n,$ $n+1$ is approximately equal to $n.$ So I answered that the sequence should convergence to $0$. Probably the interviewer was not satisfied with me. He just asked me to ponder that $\alpha$ lies between $0$ and $1$. However, it did not make any sense for me.

Am I correct? I would be happy if someone can provide me a hint for rigorous mathematical proof

(I urge the readers to mark it as duplicate if it has been already asked here)

Answer 1

By binomial expansion we have that

$$n^{\alpha}-(n+1)^{\alpha}=n^{\alpha}\left(1-\left(1+\frac1n\right)^{\alpha}\right)=$$$$=n^{\alpha}\left(1-1-\frac{\alpha}n+o\left(\frac1n\right)\right)=-\frac{\alpha}{n^{1-\alpha}}+o\left(\frac1{n^{1-\alpha}}\right)\to 0$$

Answer 2

By the mean value theorem, we have

$(n+1)^{\alpha}-n^{\alpha}=\frac{\alpha}{x_n^{1-\alpha}}$, where $x_n \in (n,n+1)$. Since $0< \alpha <1$, we have $x_n^{1-\alpha} \to \infty$, hence

$(n+1)^{\alpha}-n^{\alpha} \to 0$.

Answer 3

Another possible way is to use L'Hospital rule: $$x^\alpha - (x+1)^\alpha = \frac{1-\left(1+\frac{1}{x}\right)^\alpha}{x^{-\alpha}} \stackrel{L'Hop.}{\sim}\frac{-\alpha \left(1+\frac{1}{x}\right)^{\alpha-1}\frac{-1}{x^2}}{-\alpha x^{-\alpha-1}}= -\frac{\left(1+\frac{1}{x}\right)^{\alpha-1}}{x^{1-\alpha}}\stackrel{\alpha \in (0,1), x\rightarrow \infty}{\longrightarrow}0$$