Prove that $\frac{L(n)^m}{n}$ is a null sequence when $n \in \mathbb{N}: L(n)=k$ when $2^k\le n \le 2^{k+1}$

Question

The task is to show that

$\forall m \in \mathbb{N}: \frac{L(n)^m}{n}$ is a null sequence

$n \in \mathbb{N}: L(n)=k$ when $2^k\le n \le 2^{k+1}$

I am not allowed to use the representation of $L(n) = \lfloor\log_2{n}\rfloor$ neither am I allowed to use l'Hospital etc.

I tried to show that:

$\forall m \in \mathbb{N}: \lim\limits_{n \rightarrow \infty}{\frac{L(n)}{\sqrt[n]{m}}} = 0$ but I totally failed, because $L(n)$ is not limited.

Please give me a hint how to solve this problem.

Best Regards from Germany!

Answer 1

It is obvious that if $n\rightarrow\infty$ then $L(n)\rightarrow\infty$

From your inequality we have $2^{L(n)}\le n<2^{L(n)+1}$

So $\frac{1}{n}\le\frac{1}{2^{L(n)}}$

Then: $0<\frac{L(n)^m}{n}\le\frac{L(n)^m}{2^{L(n)}}$

And by substituting $u=L(n)$ we have $u\rightarrow\infty$ when $n\rightarrow\infty$ so:

$$\lim_{n\rightarrow\infty}\frac{L(n)^m}{2^{L(n)}}=\lim_{u\rightarrow\infty}\frac{u^m}{2^u}=0$$

So by the squeeze theorem your sequence is the null sequence.