Evaluating $\lim_{n \to +\infty} ((n+1)^a-n^a)$ I am trying to determine the limit of the following expression which arose when looking at correlations for a stochastic process.
$\lim_{n \to +\infty} ((n+1)^a-n^a)$ for values of $a\in (0,2)$.
I'm not sure how to approach this. I am aware that for $a>1$ the limit equals $\infty$ and for $a<1$ the limit is $-1$ (and $a=0 \implies$ limit $=0$).
Any ideas of what to try here would be greatly appreciated.
 A: For $a>1$ $$\lim_{x\to 0^+}\frac{(1+x)^a-1}{x^a}=\lim_{x\to0^+}\frac{1}{x^{a-1}}\cdot\frac{(1+x)^a-1}{x}=\left[\infty\cdot a\cdot(1)^{a-1}\right]=\infty$$
Therefore $\lim_{n\to\infty}(n+1)^a-n^a=\lim_{n\to\infty}\frac{\left(1+n^{-1}\right)^a-1}{n^{-a}}=\lim_{x\to0^+}\frac{(1+x)^a-1}{x^a}=\infty$
A: As I wrote in the comments, we can also use the mean-value theorem. By the MVT, there exists $x_n$ in $(n, n+1)$ such that
$$ (n+1)^\alpha - n^\alpha = \alpha x_n^{\alpha-1}. $$
Now as $x_n\in (n, n+1)$ we have $n^{\alpha-1}\leq x_{n}^{\alpha-1}\leq (n+1)^{\alpha-1}$ and hence,
$$ \alpha n^{\alpha-1}\leq (n+1)^\alpha - n^\alpha \leq \alpha (n+1)^{\alpha-1}. $$
For $\alpha \in (1,2)$ we have
$$\infty = \limsup_{n\rightarrow \infty} \alpha n^{\alpha-1} \leq \limsup_{n\rightarrow \infty} ((n+1)^\alpha - n^\alpha).$$
On the other hand, we have for $\alpha \in (0,1)$
$$ \lim_{n\rightarrow \infty} (n+1)^{\alpha-1}=0 $$
and thus by the squeeze theorem we have $\lim_{n\rightarrow \infty} ((n+1)^\alpha - n^\alpha) =0. $
A: You can use the Generalized Binomial Theorem to learn more about the behavior of the function. You may know the normal Binomial Theorem: for positive integers $n$, we have
$$(x+y)^n=\sum_{k=0}^n \binom nk x^{n-k}y^k$$
It turns out that this formula works for non-integer $n$, as long as we can generalize the binomial coefficients properly. And we do this by defining $\binom rk = \frac{(r)_k}{k!}$, where $(r)_k$ is the falling Pochhammer symbol, $(r)_k=r(r-1)\cdots(r-(k-1))$. With this, we can say that
$$(x+y)^r=\sum_{k=0}^\infty\binom rkx^{r-k}y^k$$
We can use this here to say that $(n+1)^a=\sum_{k=0}^\infty\binom ak n^{a-k}$. Keep in mind that $\binom a0=1$ for any $a$. Then we can substitute this into the limit:
$$L=\lim_{n\to\infty}\left(\sum_{k=0}^\infty\left(\binom ak n^{a-k}\right)-n^a\right)=\lim_{n\to\infty}\sum_{k=1}^\infty\binom ak n^{a-k}$$
The behavior of this limit will be dominated by the $k=1$ term, because that is the term with the largest exponent on $n$. So $L$ behaves like $\binom a1 n^{a-1}=an^{a-1}$. When $a>1$ this approaches $\infty$; for $a<1$ this will actually approach $0$, and for $a=1$ the limit also approaches $0$.
