Why didn’t we use the integral approximation (A.12) directly on \(\sum_{k = 1}^n 1/k\) to obtain an upper bound on the \(n\)-th harmonic number?.

To get an upper bound using integral approximation (A.12), we need to integrate the function from \(x = (1 - 1) = 0\). This makes the function \(\frac 1 x\) undefined because of division-by-zero. To avoid this, we took out the first term (\(k = 1\)) and carried out the sum from \(k = 2\) and the integral from \(x = 1\).