1 Example involving logarithms

Example 1.1:
Prove that \sum_{n=4}^{\infty} \frac{1}{n\ln(n)\ln(\ln(n))} diverges.
Solution.

For n\geq4, define a_n = \frac{1}{n\ln(n)\ln(\ln(n))}, and for k\geq 2, define b_k := 2^{k}a_{2^k} = \frac{2^k}{2^k\ln(2^k)\ln(\ln(2^k))}. Then, by properties of logarithms, \begin{align*} b_k = \frac{1}{k\ln(2)\ln(k\ln(2))} = \frac{1}{k\ln(2)[\ln(k) + \ln(\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.}

It turns out that this example is a special case of what is known as a generalised Bertrand series, and it’s quite surprising how general we can make this example! See here for details!