\sum_{k=1}^n\frac{1}{k(9k^2-1)} = -\sum_{k=1}^n\frac{1}{k} + \frac{3}{2}\Big(\sum_{k=1}^n\frac{1}{3k-1}+\sum_{k=1}^n\frac{1}{3k}+\sum_{k=1}^n\frac{1}{3k+1}\Big) - \sum_{k=1}^n\frac{1}{3k} = -\frac{3}{2}\sum_{k=1}^n\frac{1}{k}+\frac{3}{2}\Big(\sum_{k=1}^{3n+1}\frac{1}{k}-1\Big) = -\frac{3}{2}(\gamma_n + \ln{n}) + \frac{3}{2}(\gamma_{3n+1}+\ln{(3n+1)})-\frac{3}{2}