I have shown using the binomial expansion that

$2^n > \frac{n(n-1)(n-2)}{3!}$ for all $n \in \mathbb{Z}^+$.

My question is how can I use this fact along with the squeeze theorem to find

$\lim_{n\to \infty} \frac{n^2}{2^n}$ ?


From your identity you get, for $n\ge3$, $$ 0<\frac1{2^n}<\frac{3!}{n(n-1)(n-2)} $$ giving, for $n\ge3$, $$ 0<\frac{n^2}{2^n}<\frac{3!\cdot n^2}{n(n-1)(n-2)}=\frac{6}{(n-1)(1-\frac2n)} $$ then you can conclude as $n \to \infty$.


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