How to prove that the sequence $x_n= \frac{\ln(n)}{\sqrt{n}}$ is converging

Question

Consider the sequence $x_n = \frac{\ln(n)}{\sqrt{n}}$. It seems that the sequence tends to $0$ because $\sqrt{n}> \ln(n)$ for all $n$. I need to prove this by definition or otherwise. Can anyone give me some hints helping me proving this?

Thanks is advance.

Answer 1

For all $x>0$, $\log (1 + x) < x$. Hence, for $n\in \mathbb{N}^+$, $$ \log n = 4\log \sqrt[4]{n} \le 4(\sqrt[4]{n} - 1) < 4\sqrt[4]{n}. $$ Consequently $$ 0 \le \frac{{\log n}}{{\sqrt n }} < \frac{4}{{\sqrt[4]{n}}} \to 0. $$

Answer 2

To prove that $\sqrt n> \ln n$ for $n>4$, consider function $f(x)=\sqrt x - \ln x$. We see that $f'(x)=\frac{1}{2\sqrt x}-\frac{1}{x}=\frac{\sqrt x-2}{2x}$. Obviously, $f'(x)>0$ when $\sqrt{x}>2$ so the function is increasing and $f(4)=2-\ln 4>0$.

Answer 3

A possible approach is to use the fact (see here for a proof) $$\lim_{x\to\infty}\frac{x}{e^x}=0 $$ and consider the sequence $x_n=\ln\left(\sqrt{n}\right)\to\infty$ as $n\to\infty$. Hence $$\lim_{n\to\infty}\frac{x_n}{e^{x_n}}=\frac 12\lim_{n\to\infty}\frac{\ln(n)}{\sqrt n}=0$$

Answer 4

$x_n=2\dfrac{\log (√n)}{√n}$;

Take the limit.

Recall: $\lim_{z \rightarrow \infty} \dfrac{\log z}{z}=0.$

$\log z =\displaystyle{\int_{1}^{z}}(1/t)dt<\int_{1}^{z}(1/√t)dt$ =

$(1/2)t^{1/2}\big ]_1^z=$

$(1/2)(z^{1/2}-1) < z^{1/2}.$