Answer
$0$
Work Step by Step
Given:$a_{n}=\frac{ln n}{\sqrt n}$
A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant.
$\lim\limits_{n \to \infty}a_{n}=\lim\limits_{n \to \infty}\frac{ln n}{\sqrt n}$
Use the property: $a \ln b=\ln{b^{a}}$
$=\lim\limits_{n \to \infty}\frac{ln (n^{1/2})^{2}}{\sqrt n}$
$=\lim\limits_{n \to \infty}\frac{2 ln \sqrt n}{\sqrt n}$
Substitute $ \sqrt n=t$ and note that $t \to \infty$ when $n \to \infty$
$=\lim\limits_{t \to \infty}\frac{2 ln t}{t}$
We can also use L-Hospital's rule because the limit is of the form $\frac{\infty}{\infty}$
After applying L-Hospital's rule we get
$=\lim\limits_{t \to \infty}\frac{\frac{2}{t}}{1}$
$=\lim\limits_{t \to \infty}\frac{2}{t}$
$=0$
Because the limits exists, the sequence converges.
Hence, the given sequence converges to $0$.
