I'm looking for an analytical method for finding the smallest non-negative value of an integer $n$ such that $$\frac{n}{2^n}<\frac{1}{n^2}$$

My instinct is to manipulate the inequality into the form $$\frac{3}{\ln(2)} <\frac{n}{\ln(n)}$$but I'm now drawing an enormous blank. Is there anything I can do from here?

  • $\begingroup$ What is $n$ -- integer? real number? $\endgroup$ – user526015 Mar 26 at 15:17
  • $\begingroup$ Oops. It's an integer. $\endgroup$ – HandsomeGorilla Mar 26 at 15:31
  • $\begingroup$ There is no smallest value of $n$ as the inequality is valid for all negative $n$. $\endgroup$ – user Mar 26 at 15:34
  • $\begingroup$ You can use the LambertW function $\endgroup$ – Dr. Sonnhard Graubner Mar 26 at 15:36
  • 2
    $\begingroup$ It's a lot easier to deal with $n^3<2^n$ instead of $\tfrac3{\ln 2}< \tfrac n{\ln n}$, I think. You can find that solution mentally. $\endgroup$ – MPW Mar 26 at 15:44

Clearly the smallest integer $n>0$ at which this holds is $n=1$, but I assume you want to find the smallest $n_0>0$ such that the inequality holds for all $n \geqslant n_0$. By manipulating the expression, we equivalently want to solve for the inequality $2^n > n^3$. So we want the smallest integer $n_0$ such that $2^n > n^3$ for all $n\geqslant n_0$. The following solution does not solve for the smallest such $n_0$ by solving an algebraic equation (perhaps what the OP had in mind by "analytic") but it does provide a way to prove that such an $n_0$ works.

Consider the (differentiable) function $f(x) = 2^x - x^3$. We have $f(10) > 0$, and we claim that $f'(x) > 0$ if $x\geqslant 10$. This will show that $f(x)$ is strictly increasing for $x\geqslant 10$, so the inequality we want to show is true for all $n \geqslant n_0 = 10$, and that $10$ is the first $n_0$ for which this is true is a simple check.

Assume $x \geqslant 10$. Then, since $\ln(2)>1$ and $-3x^2 > -10x^2$, \begin{align*} f'(x) &= \ln(2)\cdot2^x-3x^2 \\ &> \ln(2)\cdot2^x-1000 \\ &>1\cdot 2^x - 1000 \\ &> 1\cdot 2^{10}-1000 \\ &= 24 > 0. \end{align*} Hence the claim.


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