\sum_{k=1}^n\frac{1}{k(4k^2-1)} = -\sum_{k=1}^n\frac{1}{k} + \sum_{k=1}^n\frac{1}{2k-1} + \sum_{k=1}^n\frac{1}{2k} + \sum_{k=1}^n\frac{1}{2k+1} - \sum_{k=1}^n\frac{1}{2k}= -\sum_{k=1}^n\frac{1}{k} + \sum_{k=1}^{2n}\frac{1}{k} - \sum_{k=1}^n\frac{1}{2k} + \sum_{k=1}^n\frac{1}{2k+1} = -\frac{3}{2}\sum_{k=1}^n\frac{1}{k} + \sum_{k=1}^{2n}\frac{1}{k} + \sum_{k=1}^n\frac{1}{2k+1} = -\frac{3}{2}\sum_{k=1}^n\frac{1}{k} + \sum_{k=1}^{2n}\frac{1}{k} + \Big(\sum_{k=1}^n\frac{1}{2k} + \sum_{k=1}^n\frac{1}{2k+1}\Big) - \sum_{k=1}^n\frac{1}{2k}