Show that $\lim_{n\to\infty}((n+1)^a - n^a) = 0$ for $0 < a < 1$ 
Show that:
  $$
\lim_{n\to\infty}((n+1)^a - n^a) = 0
$$
  where:
  $$
\begin{cases}
0 < a < 1 \\
n \in \mathbb N
\end{cases}
$$

I've started with:
$$
\begin{cases}
a = \frac{1}{1+r}\\
r\in \mathbb R \\
r > 0
\end{cases}
$$
So the expression may be rewritten as:
$$
x_n = \left(n+1\right)^{\frac{1}{1+r}} - n^{\frac{1}{1+r}} = \sqrt[{1+r}]{n+1} - \sqrt[{1+r}]{n}
$$
Looks like I need to rationalize the expression to proceed, but I'm not sure how to do it. What is the proper way of doing this? 
I guess I could use series expansion here, but that requires using derivatives which i want to keep away from for this problem.
 A: Hint Since $0 <a <1$, by the Bernoulli inequality you have 
$$(1+x)^a \leq 1+ax \qquad \forall x >0$$
Then 
$$0 \leq (n+1)^a-n^a=n^a \left( (1+\frac{1}{n})^a-1 \right)\leq n^a \left(1+\frac{a}{n}-1 \right)=\frac{a}{n^{1-a}}$$
Squeeze it.
A: By the mean value theorem,
$$
(n+1)^a-n^a = a \xi_n^{a-1}
$$
for some $\xi_n\in(n,n+1)$.
Since $a-1<0$ and $\xi_n\to\infty$, we have that $a \xi_n^{a-1}\to0$.
A: HINT
By binomial series we have that
$$(n+1)^a = n^a(1+1/n)^a=n^a(1+a/n+O(1/n^2))=n^a+\frac{a}{n^{1-a}}+O\left(\frac{1}{n^{2-a}}\right)$$
A: Without knowing $a$ it's not possible to rationalize; but you can consider instead
$$
f(t)=(1+t)^a-1
$$
and notice that
$$
\lim_{t\to0}\frac{f(t)}{t^a}=
\lim_{t\to0}\frac{f(t)}{t}t^{1-a}
$$
The limit of the first factor is $f'(0)=a$. For $0<a<1$, the second factor has limit $0$.
Now note that
$$
(1+n)^a-n^a=\frac{f(1/n)}{(1/n)^a}
$$
A: Using the binomal expansion we obtain:
$$(n+1)^a = n^a(1+1/n)^a = n^a (1+ \sum_{i=1}^a k_i \cdot 1^{i}/n^{a-i}) = n^a + n^a \sum{i=1}^a k_i 1/n^{a-i}$$
Where $k_i$ is some constant. 
Note that $1/n^{k}$ tends to zero in the limit given, so all those terms vaporize. Thus we are left with:
$$\lim_{n \to \infty} (n+1)^a - n^a = \lim_{n \to \infty} n^a - n^a = 0$$
Q.E.D.
