# Formally proving that $\lim\left[ n^2/2^n \right] = 0$ [duplicate]

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

## marked as duplicate by Jonas Meyer, GNUSupporter 8964民主女神 地下教會, Namaste, mrp, ShaileshMar 10 '17 at 15:06

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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 4:50
• Use the Binomial Theorem to show that $2^n \gt \frac{n(n-1)(n-2)}{3!}$ – André Nicolas Feb 10 '14 at 4:54

## 2 Answers

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 $n$.

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.

• sorry but if i understood correctly, when you apply "[]" greatest integer function it would be a constant sequence of zeros, right? – ronismofo Feb 10 '14 at 5:07