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.
Answer
Keep on going... the difference between the fractions is
$$\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})}$$
which, by similar reasoning as before (diff between two squares...), produces
$$\frac{-2}{(\sqrt{n-1}+\sqrt{n+1})(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})}$$
Now, as $n \to \infty$, the denominator behaves as $(2 \sqrt{n})^3 = 8 n^{3/2}$. Thus, $\lim_{n \to \infty} (-1/4) n^{-3/2} n^{2/3} = \cdots$? (Is the OP sure (s)he didn't mean $n^{3/2}$ in the numerator?)
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