calculate the the limit of the sequence $a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \sqrt{n-1} + \sqrt{n+1} -2\sqrt{n} )$ Iv'e been struggling with this one for a bit too long:
$$
a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \sqrt{n-1} + \sqrt{n+1} -2\sqrt{n} )$$
What Iv'e tried so far was using the fact that the inner expression is equivalent to that:
$$ a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \sqrt{n-1}-\sqrt{n} + \sqrt{n+1} -\sqrt{n} ) $$
Then I tried multiplying each of the expression by their conjugate and got:
$$
a_n = \lim_{n \to \infty} n^\frac{2}{3}\cdot ( \frac{1}{\sqrt{n+1} +\sqrt{n}} - \frac{1}{\sqrt{n-1} +\sqrt{n}} )
$$
But now I'm in a dead end.
Since I have this annyoing $n^\frac{2}{3}$ outside of the brackets, each of my attemps to finalize this, ends up with the undefined expression of $(\infty\cdot0)$
I've thought about using the squeeze theorem some how, but didn't manage to connect the dots right.
Thanks.

I thought I would present a way forward that is different from the approach used in the OP.  We note that from the Mean Value Theorem, there exists a number $\xi\in (x,x+1)$ and a number $\eta \in (x-1,x)$ such that 
$$f(x+1)-f(x)-f'(x)=\frac12 f''(\xi) \tag 1$$
and
$$f(x-1)-f(x)+f'(x)=\frac12 f''(\eta) \tag 2$$
Adding $(1)$ and $(2)$ yields
$$f(x+1)-2f(x)+f(x-1)=\frac12\left(f''(\xi)+f''(\eta)\right) \tag 3$$
Now, letting $f(x)=\sqrt{x}$ in $(3)$ reveals that
$$\sqrt{x+1}-2\sqrt{x}+\sqrt{x-1}=-\frac18\left(\xi^{-3/2}+\eta^{-3/2}\right)$$
Therefore, we have
$$n^{2/3}\left(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\right)=-\frac18 n^{2/3}\left(\xi^{-3/2}+\eta^{-3/2}\right)$$
since $n-1<\eta<n<\xi<n+1$.  Taking the limit as $n\to \infty$ we see that the limit of interest is $0$  
