# An Analytical method for resolving $\frac{n}{2^n} < \frac{1}{n^2}$

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?

• What is $n$ -- integer? real number? – user526015 Mar 26 at 15:17
• Oops. It's an integer. – HandsomeGorilla Mar 26 at 15:31
• There is no smallest value of $n$ as the inequality is valid for all negative $n$. – user Mar 26 at 15:34
• You can use the LambertW function – Dr. Sonnhard Graubner Mar 26 at 15:36
• 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. – 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.