I have been trying to do Problem 1.3.2 from the fantastic Almost Impossible Integrals book using different solution techniques. I tried using a Cauchy product expansion and I managed to simplify it somewhat. Provided I have not made a mistake to this point I have the result. $$ J_n = \sum_{k=2}^{\infty}\frac{2H_{k-1}}{k(k+n)} $$
The answer given in the book is: $$ J_n = \frac{2}{n}\sum_{k=1}^{n} \frac{H_k}{k} $$
If my answer is correct then we have the result: $$ \sum_{k=2}^{\infty}\frac{H_{k-1}}{k(k+n)}=\frac{1}{n}\sum_{k=1}^{n} \frac{H_k}{k} $$
Is this correct if so how can you prove it? I am positively stumped here. I would appreciate any help :)