For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one:
Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$
My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$. Note that $2^n=(1+1)^n$. Thus by the Binomial Theorem we have that, $ (1+1)^n=\sum_{k=0}^{n} {n \choose k}(1)^{n-k}(1)^n $ Evaluating some of the first couple I get the following terms: $1+n+\dfrac{n(n-1)}{2}+\dfrac{n(n-1)(n-2)}{6}+\dfrac{n(n-1)(n-2)(n-3)}{24}+...$
I have noticed that for $n<4$, $n^3 \geq 2^n$. However, it seems that when $n\geq 4, n^3 < 2^n$. Thus, would it be possible to make an argument that for $n \geq 4, s_{4}>s_{5}>s_{6}>...$? Therefore, by evaluating for $n=4$ and using some algebra you get the following:
$s_{4}=\dfrac{24n^3}{24+24n+4n(n-1)(n-2)+n(n-1)(n-2)(n-3)} \leq \dfrac{24n^3}{n^4} = \dfrac{24}{n}$
and we know that lim$_{n \rightarrow \infty} \left( \dfrac{1}{n} \right)=0$ and lim$_{n \rightarrow \infty} \left( -\dfrac{1}{n} \right)=0$
Therefore, since $\dfrac{1}{n} \geq s_{4} > s_{5} >s_{6} > ...s_{n}\geq-\dfrac{1}{n}$, by the Squeeze Theorem $s_{n}$ converges to zero as well?
Using your first idea, what about using the fact that $$(1+1)^n = \sum_{k=1}^n \binom{n}{k} \geq \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \geq \frac{(n-3)^4}{24}$$and concluding by the squeeze theorem? (as $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).