The series $\sum\limits_{n = 2}^{\infty}\frac{1}{(\ln n)^{\ln n}}$ converges. The transformation $\frac{1}{(\ln n)^{\ln n}} = \frac{1}{e^{\ln n \ln(\ln n)}}$ reveals that as \( n \) increases, \( \ln(\ln n) \) exceeds 2, leading to a rapid growth in the denominator. This growth ensures that the terms of the series approach zero sufficiently quickly, confirming convergence through the comparison test with a known convergent series.
PREREQUISITESMathematicians, students studying advanced calculus, and anyone interested in the convergence properties of series involving logarithmic terms.
Hint: $\displaystyle\frac{1}{(\ln n)^{\ln n}} = \frac{1}{e^{\ln n \ln(\ln n)}}$, and if $n$ is large enough then $\ln(\ln n)>2$.dwsmith said:$\sum\limits_{n = 2}^{\infty}\frac{1}{(\ln n)^{\ln n}}$
I am trying to show that this series diverges or converges