Deciding $\displaystyle o,\omega,\Theta$ notations

Question

I have a question which I couldn't solve for about two hours. It goes like this:

Let $\displaystyle f(n)=\left(\frac{n+3\ln(n)}{n}\right)^n \ ; \ g(n)=27^{\ln(n)}$.

Fill the blank box with the correct notation ($o,\omega,\Theta):$ $f(n)=\Box{g(n)}$.

I tried to approximate LHS to get something that is similar to RHS, but it failed.

I also tried to evaluate $\displaystyle \lim_{n\to\infty}\frac{f(n)}{g(n)}$, but it also failed.

Unfortunately, I am not familiar with other methods, so I would appreciate if you could use these.

Thanks in advance for any kind of help!

Answer 1

As $n$ is great, you have $$ \ln f(n) =n\ln\left(1+\frac{3\ln(n)}{n}\right)=n\left( \frac{3\ln n}{n}+\mathcal{O}\left( \frac{\ln^2 n}{n^2}\right)\right)=\ln (n^3)+\mathcal{O}\left( \frac{\ln^2n}{n}\right) $$ giving $$ f(n)=n^3\times e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)}. $$ On the other hand, you have $$ g(n)=27^{\ln n}=e^{(\ln n) \ln (27)}=e^{\ln (n^{\ln27})}=n^{\ln 27} $$ then $$ \frac{f(n)}{g(n)}=n^{3-\ln 27}e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)}. $$ Observe that $3-\ln 27=3-3.2958\ldots<0$, since $\displaystyle e^{\mathcal{O}\left( \frac{\ln^2n}{n}\right)} \rightarrow 1$, we deduce that $$ \frac{f(n)}{g(n)}\sim n^{3-\ln 27} \rightarrow 0. $$ We thus have $$ f(n)=o(g(n)) $$

Answer 2

As a Hint recall that $$\lim _{n \to \infty}(1+\frac{1}{n})^n=e,$$ More over the following is true $$\lim _{n \to \infty} \frac{n^3}{(1+\frac{3*ln(n)}{n})^n}=1$$ Hope this helps.