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

Do you have some tips so I could solve this problem, without the use of L'Hôpital's rule?
Indeed, we didn't see formally L'Hôpital's rule, nor Taylor series so I'm supposed to do this without such "tools".

I've tried using the fact that $n^2 = e^{2\log(n)}$ and $2^n=e^{n\log(2)}$ but didn't manage to eliminate my indeterminate form.

Thanks in advance.

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  • $\begingroup$ At least write down what you get after applying l'Hoptial's rule once. Then apply it again. $\endgroup$ – Alex R. Oct 23 '13 at 17:35
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    $\begingroup$ It look as if $n$ may range over positive integers. In that case, use the fact that by the Binomial Theorem, $(1+1)^n=1+n+\frac{n(n-1)}{2}+\frac{n(n-1)(n-2)}{6}+\cdots\gt \frac{n(n-1)(n-2)}{6}$. $\endgroup$ – André Nicolas Oct 23 '13 at 17:40
  • $\begingroup$ What, if anything, do you know about the comparative growth rates of $n$ and $\log(n)$? If you happen to know that $$\lim_{n\to\infty}\bigl[n-\log(n)\bigr]=\infty,$$ then since $$\frac{n^2}{2^n}=\frac{e^{2\log(n)}}{e^{n\log(2)}}=e^{2\log(n)-n\log(2)},$$ and since $\log(2)>0,$ it follows that the limit you're looking for is $0$, and even gives you a start on how it can be proven. $\endgroup$ – Cameron Buie Oct 23 '13 at 17:53
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There are many ways to do that. For example, you can prove that for $n \ge 10$ we have $$ 2^n > n^3. $$ You can show that using the induction argument.

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Regarding to analysis tag, you seem to be familiar to series so let us consider the following series: $$\sum(n^2/2^n)$$ it is not hard seeing that this series is convergent and so the $n-$th term approaches to zero while $n\to\infty$. Clearly, this way needn't using L'Hopital's rule.

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  • $\begingroup$ @Amzoti: Thanks for your support my dear friend. I hope you have a good time ahead and a peaceful slumber. :-) $\endgroup$ – mrs Oct 24 '13 at 1:48
  • $\begingroup$ You are welcome my friend! I appreciate all of the hard work you and amWhy put into the site helping people become better problem solvers! Time to watch a movie and relax as it has been a difficult month at work and preparing two talks! Have a great night! Regards $\endgroup$ – Amzoti Oct 24 '13 at 1:49
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I noticed the OP had wanted L'Hôpital's rule off the table only after I posted my previous reply. I apologize for my inattention. Here is another answer:

Let $a_n=n^2/2^n$ for all $n\in\mathbb{Z}_+$. You can easily compute that \begin{align*} a_{n+1}-a_n=-\frac{n(n-2)-1}{2^{n+1}}, \end{align*} which is negative for $n\geq3$. Therefore, the sequence is eventually decreasing and bounded below (by zero), so it must have a limit.

Obviously, this limit cannot be negative, since $a_n>0$ for all $n$. If we show it cannot be positive, either, then we can conclude that the limit must be zero.

Suppose that the limit is positive: $\lim_{n\to\infty} a_n=c>0$. Since the sequence is decreasing (after omitting $a_1$ and $a_2$, which does not affect convergence properties), it is also true that \begin{align*} c=\lim_{n\to\infty} a_n=\inf_{n\in\mathbb{Z}_+\setminus\{1,2\}}a_n. \end{align*} Then, for $n$ large enough and beyond, we have that \begin{align*} (*)\quad c\leq a_n\leq\frac{3}{2}c. \end{align*}

Therefore, \begin{align*} (**)\quad a_{n+1}=a_{n}\times\frac{a_{n+1}}{a_n}=a_n\times\frac{(n+1)^2}{2n^2}\leq\frac{3}{2}c\times \frac{(n+1)^2}{2n^2}, \end{align*}

where the last inequality follows from $(*)$. But $(n+1)^2/(2 n^2)$ converges to $1/2$ as $n\to\infty$, so that for $n$ sufficiently large and beyond, \begin{align*} \frac{(n+1)^2}{2n^2}\leq\frac{5}{8}. \end{align*} For such values of $n$, $(**)$ implies that \begin{align*} a_{n+1}\leq\frac{3}{2}c\times\frac{(n+1)^2}{2n^2}\leq\frac{15}{16}c<c, \end{align*} where the last inequality is strict because $c$ is positive. But this contradicts $c$ being the infimum of the values of the sequence. The limit thus cannot be positive.

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  • $\begingroup$ +1. I appreciate your effort of writing another answer after I pointed out that L'H was off-limits. :) $\endgroup$ – Lord_Farin Oct 23 '13 at 20:13
  • $\begingroup$ Thanks, @Lord_Farin. :-) $\endgroup$ – triple_sec Oct 23 '13 at 20:37

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