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Not sure how to formally prove this (specifically regarding the choice of $\epsilon$ in the formal limit definition)... Any suggestions?

michael straws
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  • To prove that $;n^2/2^n < 1;$ for all $;n>N;$ , for some specific $;N\in\Bbb N;$ , is formal enough. Why do you want to mess around with $;\epsilon ;$ and stuff? – DonAntonio Feb 10 '14 at 04:50
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    Use the Binomial Theorem to show that $2^n \gt \frac{n(n-1)(n-2)}{3!}$ – André Nicolas Feb 10 '14 at 04:54

2 Answers2

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Sketch: Show that $2^n \ge n^3$ for sufficiently large $n$ (easily done by induction). Then you have

$$\left|\frac{n^2}{2^n} - 0\right| \le \frac{n^2}{n^3} = \frac 1 n$$

Now studying the convergence is much easier, since $1/n < 1$ for such $n$.

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I hope you can prove $n^2<2^n$ by induction if $n\geq 4$ and then you surely have

the inequality $\displaystyle {0<\frac {n^2}{2^n}<1}$ boils down to your conclusion.

Maximal_inequality
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