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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?

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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 $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).

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  • $\begingroup$ I'm pretty sure this is what I was looking for. I think you pretty much condensed what i was trying to say; all of this has just been my scratch work and ideas. Thank you! $\endgroup$ – FlashKicks909 May 5 '16 at 18:38
  • $\begingroup$ You're welcome! $\endgroup$ – Clement C. May 5 '16 at 18:54
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Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.

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  • $\begingroup$ I'm taking an introductory analysis class. I can see why that is true; however, we would have to prove it and I don't want to do another induction proof. $\endgroup$ – FlashKicks909 May 5 '16 at 18:37
  • $\begingroup$ @Kf-Sansoo, if I wanted to prove your "trick", then how formal would saying "because exponential grows faster than any polynomial" be ? $\endgroup$ – Aritra Das May 5 '16 at 18:49
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    $\begingroup$ 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. $\endgroup$ – Mark Viola May 5 '16 at 18:57
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    $\begingroup$ @Kf-Sansoo You wrote "Your comment showed you dont yet know real analysis" in response to Dr. MV. You ought to think twice before making such statements. In this case it wasn't just rude, it was false. $\endgroup$ – zhw. May 6 '16 at 0:01

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