Asymptotic equality proof with $a_n^2 \ln a_n ~ n$

Question

Given $a_n^2 \ln a_n \sim n$, prove that $a_n \sim \sqrt{\frac{2n}{\ln n}}$.

How do I approach this?

Answer 1

$$a_n^2\ln(a_n)=n(1+o(1))$$ $$\ln(a_n^2\ln(a_n))=\ln(n(1+o(1))=\ln(n)+\ln(1+o(1))=\ln(n)+o(1)$$ $$2\ln(a_n)+\ln(\ln(a_n))=\ln(n)+o(1)$$ Now by the first asymptotic we can conclude $a_n$ increases without bound as $n$ increases so,

$$\lim_{n\to \infty}\frac{\ln(\ln(a_n))}{\ln(a_n)}=0$$

Now divide both sides by $\ln(a_n)$ in our last equality, and simplify to obtain

$$2+o(1)=\frac{\ln(n)}{\ln(a_n)}$$

Multiply both sides by $\ln(a_n)$ to obtain,

$$2\ln(a_n)+o(\ln(a_n))=\ln(n)$$

Now multiply both sides by $a_n^2$

$$2a_n^2\ln(a_n)+o(a_n^2\ln(a_n))=a_n^2\ln(n)$$

Now sense $a_n^2\ln(a_n)=n+o(n)$, we get that

$$2n+o(n)=a_n^2\ln(n)$$

$$\frac{2n}{\ln(n)}+o(\frac{n}{\ln(n)})=a_n^2$$

And thus,

$$a_n=\sqrt{\frac{2n}{\ln(n)}}+o(\sqrt{\frac{n}{\ln(n)}})$$

Or restated in your form, $$a_n\sim \sqrt{\frac{2n}{\ln(n)}}$$