Limit of sequence $\lim \limits_{n \to \infty\ }(\ln n)^{1+\frac{(\ln(\ln n))^{2019}}{\ln n}}-\frac{\sqrt[2018]n}{\ln n}$

Question

$\lim \limits_{n \to \infty\ }(\ln n)^{1+\frac{(\ln(\ln n))^{2019}}{\ln n}}-\frac{\sqrt[2018]n}{\ln n}$

I can show that $\lim \limits_{n \to \infty\ } {\frac{(\ln(\ln n))^{2019}}{\ln n}}=0 $, so $\lim \limits_{n \to \infty\ }(\ln n)^{1+\frac{(\ln(\ln n))^{2019}}{\ln n}}=+\infty$

and $\lim \limits_{n \to \infty\ } \frac{\sqrt[2018]n}{\ln n}=-\infty$

But I don't know how take it together, and show that $\lim \limits_{n \to \infty\ }a_n=-\infty$.

Could someone give me hints?

Answer 1

We have that

$$(\ln n)^{1+\frac{(\ln(\ln n))^{2019}}{\ln n}}=\ln n \cdot e^{\frac{(\ln(\ln n))^{2020}}{\ln n}}$$

and then

$$(\ln n)^{1+\frac{(\ln(\ln n))^{2019}}{\ln n}}-\frac{\sqrt[2018]n}{\ln n}= \ln n \cdot e^{\frac{(\ln(\ln n))^{2020}}{\ln n}}-\frac{\sqrt[2018]n}{\ln n}=\ln n \left(e^{\frac{(\ln(\ln n))^{2020}}{\ln n}}-\frac{\sqrt[2018]n}{\ln^2 n}\right)\to-\infty$$

indeed $\forall a>0$

• $\frac{n^a}{\log n} \to \infty$
• $\frac{(\ln(\ln n))^{a}}{\ln n} \to 0$