# Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$

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 $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).
Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.
• This inequality is not even valid for all $n\ge 1$. While it is true for "sufficiently large $n$," taking this as a given seems tantamount to taking the limit of interest as given. – Mark Viola May 5 '16 at 18:57