Determine the convergence or divergence of the sequence with the given \( n \)th term. If the sequence converges, find its limit. (If the quantity diverges, enter DIVERGES.) \[ a_{n}=\frac{\ln \left(n^{9}\right)}{4 n} \]

To analyze the sequence \( a_{n} = \frac{\ln(n^{9})}{4n} \), we can simplify it: \[ a_{n} = \frac{9 \ln(n)}{4n}. \] As \( n \to \infty \), we can apply L'Hôpital's Rule to the form \( \frac{\ln(n)}{n} \) since both the numerator and denominator approach infinity. Differentiating the numerator and the denominator gives: 1. Derivative of the numerator: \( \frac{d}{dn}[\ln(n)] = \frac{1}{n} \). 2. Derivative of the denominator: \( \frac{d}{dn}[n] = 1 \). Applying L'Hôpital's Rule: \[ \lim_{n \to \infty} \frac{\ln(n)}{n} = \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0. \] Now we can find our limit: \[ \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{9 \ln(n)}{4n} = \frac{9}{4} \cdot \lim_{n \to \infty} \frac{\ln(n)}{n} = \frac{9}{4} \cdot 0 = 0. \] Thus, the sequence converges, and the limit is: \[ \lim_{n \to \infty} a_{n} = 0. \]