How does $H_{k}=\displaystyle\left[H_{k+1}-\frac{1}{k+1}\right]$?

Question

I'm confused about one of the algebraic steps,

In showing the $k+1$'th term, we have:

\begin{align}\displaystyle\sum\limits_{j=1}^{k+1}H_{j} &= \displaystyle\sum\limits_{j=1}^{k}H_{j} + H_{k+1} \\ &=(k+1)H_{k}-k+H_{k+1}\\ &=(k+1)\left[H_{k+1}-\frac{1}{k+1}\right]-k+H_{k+1} <----\text{this step} \\ &=\vdots\\ &=(k+2)H_{k+1}-(k+1). \end{align}

How does $H_{k}=\displaystyle\left[H_{k+1}-\frac{1}{k+1}\right]$?

i.e. for $k=5$, $H_{5}=\displaystyle\frac{1}{5}$, but $H_{5+1}-\displaystyle\frac{1}{5+1}=\frac{1}{6}-\frac{1}{6}=0.$

Answer 1

Observe that $$ \color{blue}{H_{k}=1+\frac12+\frac13+\cdots+\frac 1 k} $$ then $$ H_{k+1}=\color{blue}{1+\frac12+\frac13+\cdots+\frac 1 k}+\color{red}{\frac 1{k+1}} $$ thus $$ H_{k+1}=\color{blue}{H_k}+\color{red}{\frac 1{k+1}} $$ equivalently $$ \color{blue}{H_k}=H_{k+1}-\color{red}{\frac 1{k+1}} $$