I want to a know tight bound of this quantity when $n$ is even $$\sum_{k=1}^{n/2}\sum_{m=n-k}^{n}\frac{1}{k(k+1)m}$$. I simplified the expression and it comes like $$H_n[1-\frac{1}{n/2+1}]-\sum_{k=1}^{n/2}\frac{H_{n-k}}{k} + \sum_{k=1}^{n/2}\frac{H_{n-k}}{k+1}$$ where $H_n$ is the n-th partial sum of the harmonic series.
Not able to make much progress after this.