limit with exponential

Question

I am trying to solve asymptotic relation between 2 functions: $$f(n)=2^n*n$$ $$g(n)=\frac {3^n}{n^2} $$

I started to solve $$\lim_{x\to \infty} \frac{2^n*n^3}{3^n}=\lim_{x\to \infty} (\frac {2}{3})^n*n^3$$

Than I try to use L'Hopital but it doesn't work.

So I try to solve inequality:

$$f(n) <= g(n)$$ $$2^n*n <= c*\frac {3^n}{n^2}$$ $$2^n*n^3 <= c*3^n$$ $$e^{n*\ln 2}*n^3 <= c*e^{n*\ln 3}$$ $$n^3 <= c$$

And this means that $f(n) > c*g(n)$. But according to WolframAlpha limit is $0$ which means that result is inverse $f(n)=o(g(n))$.

Please give me a hint to solve it. Thank you.

Answer 1

$$\frac{f(n)}{g(n)}=\frac{2^nn^3}{3^n}$$

If we now apply the $\;n-$th root test to the corresponding series, we get:

$$\sqrt[n]{\frac{f(n)}{g(n)}}=\frac23\sqrt[n]{n^3}\xrightarrow[n\to\infty]{}\frac23<1\implies \;\text{the series converges}\implies$$

$$\frac{f(n)}{g(n)}\xrightarrow[n\to\infty]{}0$$

Answer 2

The elementary way taught in first semester calculus is to write $|q|^{-1}=1+\alpha$, where here with $q=\frac23$ we get $α=\frac12$, and use the binomial theorem to show that

$$q^{-n}=(1+α)^n=1+nα+\tbinom n2 α^2+\dots+α^n\ge 1+nα+\tbinom n2 α^2,$$

so that

$$n\cdot |q|^n\le \frac{n}{1+nα+\frac{n(n-1)}2 α^2}$$

where the bound on the right side clearly converges to zero.