Show that: $\log^{2019} n = o(n)$

Question

That is to find $a,\;c$ and $n$ such that: $\log^{2019} n = o(n)$

Answer 1

Observe below that for all $n>1$, $$n=n$$ $$n^\frac{1}{2019}=n^\frac{1}{2019}$$ $$\log n^\frac{1}{2019} <n^\frac{1}{2019}$$ $$\frac{1}{2019} \log n <n^\frac{1}{2019}$$ $$\bigg( \frac{1}{2019} \log n \bigg) ^{2019} < (n^\frac{1}{2019})^{2019}$$ $$\bigg( \frac{1}{2019} \bigg)^{2019}( \log n ) ^{2019} < n$$ $$\frac{1}{2019^{2019}}( \log n ) ^{2019} < n$$ $$( \log n ) ^{2019} < 2019^{2019}n$$ $$ \log^{2019} n < 2019^{2019}n$$

where $2019^{2019} = c \in \mathbb{R}$. Thus, by substitution we have

$$\log^{2019} n < c \cdot n$$

Answer 2

For any $c > 0$ and $n > 1$,

$\begin{array}\\ \log(n) &=\int_1^n \dfrac{dt}{t}\\ &=\int_1^n t^{-1}dt\\ &<\int_1^n t^{-1+c}dt\\ &=\dfrac{t^c}{c}|_1^n\\ &=\dfrac{n^c-1}{c}\\ &<\dfrac{n^c}{c}\\ \end{array} $

Letting $c = a/2$, $\log(n) \lt \dfrac{n^{a/2}}{a/2} = \dfrac{2n^{a/2}}{a} = n^a\dfrac{2n^{-a/2}}{a} $ for any $a > 0$.

Putting $a = 1/m$,

$\begin{array}\\ \log(n) &\lt n^{1/m}(m2n^{-1/(2m)})\\ \text{so}\\ \log^m(n) &\lt n(m2n^{-1/(2m)})^m\\ &= n((2m)^mn^{-1/2}) \qquad\text{since } (2m)^mn^{-1/2} \to 0\\ &\in o(n)\\ \end{array} $

Answer 3

We can see$$\lim_{x\to\infty}\frac{\log x}{x^{\frac{1}{2019}}}=\lim_{x\to\infty}\frac{\frac1x}{\frac{1}{2019}x^{\frac{1}{2019}-1}}=0$$ so$$\lim_{x\to\infty}\frac{\log^{2019} x}{x}=0$$