Show that \(\sum_{k=1}^{n} 1/(2k - 1) = \ln(\sqrt{n}) + O(1)\) by manipulating the harmonic series.

Observe that the summation \(\sum_{k=1}^{n} 1/(2k - 1)\) is a harmonic series from \(1\) to \(2n\) with all terms that have even denominators removed. With this in mind, we can express this summation as a function of modified harmonic series’ in order to solve using the harmonic series identity: \(\sum\limits_{k=1}^{n}\frac{1}{k} = \ln n + O(1)\).

\[\begin{split} \sum\limits_{k=1}^{n} \frac{1}{2k - 1} & = 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n -1 }\\ & = \sum\limits_{k=1}^{2n}\frac{1}{k} - \sum\limits_{k=1}^{n}\frac{1}{2k} \\ & = \sum\limits_{k=1}^{2n}\frac{1}{k} - \frac{1}{2} \sum\limits_{k=1}^{n}\frac{1}{k} \\ & = \ln (2n) + O(1) - \frac{1}{2}(\ln n + O(1)) \\ & = \ln n + \ln 2 + O(1) - \frac{1}{2} \ln n - \frac{1}{2} O(1) \\ & = \frac{1}{2} \ln n + O(1) \\ & = \ln (\sqrt{n}) + O(1) \\ \end{split}\]