$\lim_{n \to \infty} \frac{\ln x^q}{x^p}$ not necessarily =0 for any $p>0$ and $q>0$ right?

Question

Is it true or false that$$\lim_{n \to \infty} \frac{\ln n^q}{n^p}=0$$ is not necessarily true for all $p>0$ and $q>0$?

I understand that $$\lim_{n \to \infty} \frac{(\ln n)^q}{n^p}=0$$ for all $p>0$ and $q>0$.

Answer 1

Assuming you mean: $$ \lim_{n \to \infty} \frac{\ln n^q}{n^p} = \lim_{n \to \infty} \frac{q \ln n}{n^p} = 0 $$ This for all $p > 0$

Answer 2

Besides to other answers, note that the series : $$\sum_{n=1}^{\infty}\frac{(\ln n)^q}{n^p}$$ is convergent when $p>1, q\leq0$, since $$\lim_{n\to\infty}n^p\frac{(\ln n)^q}{n^p}<\infty$$

Answer 3

I understand that $$\lim_{n \to \infty} \frac{(\ln n)^q}{x^p}=0\tag{1}$$ for all $p>0$ and $q>0$.

If you know this to be true, then we can apply it here: for all $p, q > 0$ $$ \begin{align}\lim_{n \to \infty} \frac{\ln n^q}{x^p} & = \lim_{n \to \infty} \frac {q\,\ln n}{n^p}\\ \\ & = q\lim_{n \to \infty} \frac {(\ln n)^1}{n^p} \tag{2} \\ \\ & = q\cdot 0 \tag{By (1), $q = 1 \implies (2) = q\cdot 0)$ } \\ \\ & = 0 \\ \\ \end{align}$$

Answer 4

Edit: well finally it was $x=n$ and not $n=x$. The following is still true and it obviously implies the $n$ case. So I'll leave it like this.

The first ($\forall p,q>0$) follows from the second (with $q=1$), as $\ln x^q=q\ln x$. So let us prove the second. I'll do the general case for the sake of completeness, although we only need $q=1$.

Set $u:=x^p$, i.e. $x=u^{1/p}$. Then $$ \frac{(\ln x)^q}{x^p}=\left(\frac{1}{p}\right)^q\frac{(\log u)^q}{u}. $$ Applying L'Hospital $n$ times for some $n>q$ (e.g. $n=\lfloor q\rfloor+1$) yields $$ \lim_{+\infty}\frac{(\log u)^q}{u}=q\lim_{+\infty}\frac{(\log u)^{q-1}}{u}=\ldots=q(q-1)\cdots(q-n+1)\lim_{+\infty}\frac{1}{u(\log u)^{n-q}}=0. $$ Since $u\longrightarrow+\infty$ if and only if $x\longrightarrow +\infty$, you are done proving $$ \lim_{+\infty}\frac{(\ln x)^q}{x^p}=0\qquad\forall p>0,q>0. $$