MathItems

Proof P6 of T3

k=1nHk=k=1n1Hk+Hn=nHnn+Hn=(n+1)Hnn\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}

where the second equality sign uses T4.