2 Another example involving logarithms
For \(n\geq3\), define \[a_n = \frac{n^2}{\left(\ln(\ln(n))\right)^{\ln(n)}},\] and for \(k\geq 2\), define \[b_k := 2^{k}a_{2^k} = \frac{2^{3k}}{\left(\ln(\ln(2^k))\right)^{\ln(2^k)}}.\] Then, by properties of logarithms, \[\begin{align*} b_k = \frac{2^{3k}}{\left(\ln(k\ln(2))\right)^{k\ln(2)}} = \frac{2^{3k}}{[\ln(k) + \ln(\ln(2))]^{k\ln(2)}}. \end{align*}\] Also, for \(k\geq 2\), we know that (under the assumption that \(\ln : (0,\infty) \to \mathbb{R}\) is an increasing function), \[\ln(k) \geq \ln(2) \geq \ln(\ln(2))).\] Hence \[b_k \geq \frac{1}{2\ln(2)k\ln(k)} =: c_k.\] From lectures (or by applying the Cauchy condensation test again to \(c_k\)), we know that \(\sum_{k=2}^{\infty}c_k\) diverges. Hence, by comparison (as \(c_k \geq 0\)), we find that \(\sum_{k=2}^{\infty} b_k\) diverges. Finally, by the Cauchy condensation test, we conclude that \[\sum_{n=4}^{\infty} a_n = \sum_{n=4}^{\infty} \frac{1}{n\ln(n)\ln(\ln(n))} \quad \text{diverges.}\]