The limit of $(n+1)^k - n^k$ when $0<k<1$

Question

I'm trying to prove that for any given real number $0<k<1$, $$\lim_{n\to \infty} (n+1)^k - n^k=0$$

It is easy to prove that if $k=\frac1m$ for some positive integer $m$, $$(n+1)^\frac 1m - n^\frac 1m\to0\quad \text{as }n\to\infty$$
simply using the following formula. $$(a-b)=(a^\frac 1m - b^\frac1m)(a^\frac{m-1}m + a^\frac{m-2}m b^\frac 1m+....+b^\frac {m-1} m)$$ And letting $a=\alpha^N$, $b=\beta^N$ the limit $\lim_{n\to \infty} (n+1)^k - n^k$ goes $0$ for any rational number $0<k=\frac Nm<1$.
However, for any given real number, $\pi$ to say, how can I even evaluate limit like this? $$\lim_{n\to \infty} (n+1)^\frac 1\pi - n^\frac 1\pi$$
I don't know how to deal with irrational numbers at all. Any advice would be a great help. Thank you in advance.

Answer 1

Because $$(n+1)^k-n^k=n^k\left(\left(1+\frac{1}{n}\right)^k-1\right)<n^k\left(\left(1+\frac{1}{n}\right)-1\right)=\frac{1}{n^{1-k}}\rightarrow0.$$

Answer 2

Notice that the function $\alpha \mapsto (n+1)^\alpha - n^\alpha$ is increasing, which you can check by taking the derivative.

For any $0 < \alpha < 1$ pick $k \in \mathbb{N}$ such that $\frac1k \le \alpha < \frac1{k+1}$.

Then we have

$$(n+1)^{\frac1{k+1}} - n^{\frac1{k+1}} \le (n+1)^{\alpha} - n^{\alpha} \le (n+1)^{\frac1k} - n^{\frac1k} $$

Letting $n\to\infty$ we get $\lim_{n\to\infty} (n+1)^{\alpha} - n^{\alpha} = 0$ by the squeeze theorem.

Answer 3

MVT:

$\dfrac{(n+1)^k-n^k}{1}=k(t^{k-1})$, where $t \in (n,n+1)$.

$0 < (n+1)^k-n^k =$

$=k\dfrac{1}{t^{1-k}} \lt k\dfrac{1}{n^{1-k}}.$

Take the limit using squeeze theorem.