Proof P6 of T3 ∑k=1nHk=∑k=1n−1Hk+Hn=nHn−n+Hn=(n+1)Hn−n\begin{aligned} \sum_{k=1}^nH_k&=\sum_{k=1}^{n-1}H_k+H_n\\ &=nH_n-n+H_n\\ &=(n+1)H_n-n \end{aligned}k=1∑nHk=k=1∑n−1Hk+Hn=nHn−n+Hn=(n+1)Hn−n where the second equality sign uses T4.