We are given the expression $a_n = \frac{\ln(\frac{1}{n})}{\sqrt{2n}}$ and we need to analyze it. The question is implicitly asking about the limit of $a_n$ as $n$ approaches infinity.

1. Problem Description

We are given the expression $a_n = \frac{\ln(\frac{1}{n})}{\sqrt{2n}}$ and we need to analyze it. The question is implicitly asking about the limit of $a_n$ as $n$ approaches infinity.

2. Solution Steps

We have the expression $a_n = \frac{\ln(\frac{1}{n})}{\sqrt{2n}}$.

Using the property of logarithms, $\ln(\frac{1}{n}) = \ln(1) - \ln(n) = -\ln(n)$.

Therefore, $a_n = \frac{-\ln(n)}{\sqrt{2n}}$.

We want to find the limit of $a_n$ as $n \to \infty$, i.e., $\lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2n}}$.

We can rewrite this as $\lim_{n \to \infty} \frac{-\ln(n)}{\sqrt{2} \sqrt{n}}$.

This is of the form $\frac{-\infty}{\infty}$, so we can apply L'Hopital's rule.

Taking the derivative of the numerator with respect to $n$, we have $\frac{d}{dn}(-\ln(n)) = -\frac{1}{n}$.

Taking the derivative of the denominator with respect to $n$, we have $\frac{d}{dn}(\sqrt{2n}) = \sqrt{2} \frac{d}{dn}(\sqrt{n}) = \sqrt{2} \cdot \frac{1}{2\sqrt{n}} = \frac{\sqrt{2}}{2\sqrt{n}}$.

Applying L'Hopital's rule, we get:

$\lim_{n \to \infty} \frac{-\frac{1}{n}}{\frac{\sqrt{2}}{2\sqrt{n}}} = \lim_{n \to \infty} \frac{-1}{n} \cdot \frac{2\sqrt{n}}{\sqrt{2}} = \lim_{n \to \infty} \frac{-2\sqrt{n}}{\sqrt{2}n} = \lim_{n \to \infty} \frac{-2}{\sqrt{2}\sqrt{n}} = \lim_{n \to \infty} \frac{-\sqrt{2}}{\sqrt{n}}$.

As $n \to \infty$, $\sqrt{n} \to \infty$, so $\frac{-\sqrt{2}}{\sqrt{n}} \to 0$.

Therefore, the limit is

0.

3. Final Answer

$\lim_{n \to \infty} a_n = 0$