1
$\begingroup$

This question already has an answer here:

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

$\endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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 $n$.

$\endgroup$
0
$\begingroup$

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.