Question: I would like to prove that $$\lim\_{n\to\infty} \sqrt[3]{(n+1)(n+2)(n+3)}-n = 2,$$
but have no ideas on how to proceed with this limit. Any hint? Thanks in advance

Answer: 
Hint : factorize by $n$ and use a Taylor expansion of $f(t)=(1+t)^\frac{1}{3}$ when $t$ is near zero.
$$\lim\_{n\to\infty} \sqrt[3]{(n+1)(n+2)(n+3)}-n$$

Using $\bf{A.M\geq G.M}\;,$

$$\lim\_{n\rightarrow \infty}\frac{(n+1)+(n+2)+(n+3)}{3}\geq \lim\_{n\rightarrow \infty}\sqrt[3]{(n+1)(n+2)(n+3)}$$

So we get $$\lim\_{n\rightarrow \infty} n+2\geq\lim\_{n\rightarrow \infty} \sqrt[3]{(n+1)(n+2)(n+3)}$$

and equality will hold when, $$\lim\_{n\rightarrow \infty}(n+1)=\lim\_{n\rightarrow \infty}(n+2) = \lim\_{n\rightarrow \infty}(n+3)$$

So we get $$\lim\_{n\to\infty} \sqrt[3]{(n+1)(n+2)(n+3)}-n = 2$$