Evaluate $\lim_{n\to \infty} \dfrac{\ln(n)^n}{5n^n}$

Question

Evaluate $$\lim_{n\to \infty} \frac{\ln(n)^n}{5n^n}$$

$$\lim_{n\to \infty} \frac{\ln(n)^n}{5n^n}=\lim_{n\to \infty} \frac{n\cdot \ln(n)}{5^nn^n}=\lim_{n\to \infty} \frac{\ln(n)}{5^nn^{n-1}}$$

$\ln(n)<<5^nn^{n-1}$ and therefore $$\lim_{n\to \infty} \frac{\ln(n)^n}{5n^n}=0$$

Is there a way to solve it without L'Hospital? and $\ln(n)<<5^nn^{n-1}$?

Answer 1

\begin{align*} \log n=\log n-\log 1=\int_{1}^{n}\dfrac{1}{t}dt\leq\int_{1}^{n}\dfrac{1}{t^{1/2}}dt=2n^{1/2}-2, \end{align*} so \begin{align*} \dfrac{\log(n^{n})}{5n^{n}}\leq\dfrac{2n^{n/2}-2}{5n^{n}}=\dfrac{2}{5}\dfrac{1}{n^{n/2}}-\dfrac{2}{5}\dfrac{1}{n^{n}}\rightarrow 0. \end{align*}

Answer 2

Note that

$$\frac{\ln(n)^n}{5n^n}=\frac{n\ln n }{5n^n}=\frac{\ln n }{5n^{n-1}}\le\frac{\ln n }{n}\to0$$

thus for squeeze theorem

$$\lim_{n\to \infty} \frac{\ln n^n}{5n^n}=0$$

Answer 3

Let $n \in \mathbb {Z^+}$:

$\log(n) =$

$\displaystyle \int_{1}^{n} \dfrac{1}{x}dx \lt 1×n $ (upper sum).

Hence:

$\dfrac{\log(n)}{n} \lt 1.$

$\dfrac{n\log(n)}{5n^n} \lt \dfrac{1}{5n^{n-2}}.$

The limit $n \rightarrow \infty$ is?