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.

Consider the corresponding function: if it has a limit at $\infty$, it will be the same as for the sequence. For the limit of the function we can do a substitution $x=1/t$, so we have to compute
\begin{align}
\lim_{t\to0^+}
\frac{1}{t^{2/3}}\frac{\sqrt{1-t}+\sqrt{1+t}-2}{\sqrt{t}}
&=
\lim_{t\to0^+}\frac{1-t/2-t^2/8+o(t^2)+1+t/2-t^2/8+o(t^2)-2}{t^{7/6}}
\\
&=\lim_{t\to0^+}\frac{-t^2/4+o(t^2)}{t^{7/6}}=0
\end{align}
If the exponent in the first factor is $3/2$, rather than $2/3$, the denominator would be $t^2$ and the limit would be $-1/4$.
