\( \sum _ { n = 1 } ^ { \infty } \frac { n ^ { n } } { ( \ln ( n ) ) ^ { n } } \)

To analyze the convergence of the series \( \sum _ { n = 1 } ^ { \infty } \frac { n ^ { n } } { ( \ln ( n ) ) ^ { n } } \), we can apply the ratio test. We explore the behavior of the terms as \( n \) approaches infinity. Consider the \( n \)-th term of the series: \[ a_n = \frac{n^n}{(\ln n)^n}. \] Now, evaluate the ratio of successive terms: \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(\ln(n+1))^{n+1}} \cdot \frac{(\ln n)^n}{n^n}. \] This simplifies to: \[ \frac{(n+1)^{n+1}}{n^n} \cdot \frac{(\ln n)^n}{(\ln(n+1))^{n+1}}. \] We can simplify it further: \[ \frac{(n+1)^{n+1}}{n^n} = (n+1)\left(\frac{n+1}{n}\right)^n = (n+1)\left(1 + \frac{1}{n}\right)^n \sim (n+1)e \text{ as } n \to \infty. \] The logarithmic term can be approximated using \( \ln(n+1) \sim \ln n + \frac{1}{n} \) for large \( n \): \[ \frac{(\ln n)^n}{(\ln(n+1))^{n+1}} \approx \frac{(\ln n)^n}{(\ln n + \frac{1}{n})^{n+1}} \sim \frac{(\ln n)^n}{(\ln n)^{n+1}} \text{ as } n \to \infty, \] which approaches \( \frac{1}{\ln n} \). Thus, we get that: \[ \frac{a_{n+1}}{a_n} \sim (n+1)e \cdot \frac{1}{\ln n}. \] As \( n \to \infty \), the ratio diverges since \( e(n + 1)/\ln n \to \infty \). Therefore, by the ratio test, the series diverges. In conclusion: \[ \sum _ { n = 1 } ^ { \infty } \frac { n ^ { n } } { ( \ln ( n ) ) ^ { n } } \text{ diverges.} \]