Proof P5 of T4 ∑k=1n−1Hk=∑k=1nHk−Hn=(n+1)Hn−n−Hn=nHn−n\begin{aligned} \sum_{k=1}^{n-1}H_k&=\sum_{k=1}^nH_k-H_n\\ &=(n+1)H_n-n-H_n\\ &=nH_n-n \end{aligned}k=1∑n−1Hk=k=1∑nHk−Hn=(n+1)Hn−n−Hn=nHn−n where the second equality sign uses T3.