Rewriting an infinite series

Question

I was reading this economics article (page 2), and I was curious about how to go from $$ DPV \space w(1)=w(1)+\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+...+\frac{w(1)}{(1+r)^\infty }$$ to this part $$DPV\space [w(1)\times\frac{1}{1+r}]=\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+...+\frac{w(1)}{(1+r)^\infty }$$ $$DPV\space [w(1)\times\frac{1}{1+r}]=w(1)$$ $$DPV\space [w(1)\times (1-\frac{1}{1+r})]=w(1)$$ $$DPV\space w(1)=w(1)\frac{r+1}{r} $$

(DPV is the discounted present value, and $w(1)$ is the wage after 1 year of education).

Answer 1

Something useful to know is that $d=u (1+r)^{-1} $ grows to $u $ after one period since $d (1+r)=u $. So $d $ is the duscounted present value of $u$.

The sum -DPV of a perpetuity of $w (1) $- is a geometric series and sums to $w (1)(1+r)/r $).

Answer 2

$$ DPV \space w(1)=w(1)+\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+\cdots+\frac{w(1)}{(1+r)^\infty }\tag 1$$ Multiplying the LHS and the RHS by $\frac{1}{1+r}$ we have to this part $$ \begin{align} DPV\space w(1)\times\color{red}{\frac{1}{1+r}}&=\color{red}{\frac{1}{1+r}}\times\left[w(1)+\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+\cdots+\frac{w(1)}{(1+r)^\infty }\right]\\ &=\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+\frac{w(1)}{(1+r)^3 }+\cdots+\frac{w(1)}{(1+r)^\infty }\tag 2 \end{align} $$ Subtracting $(1)$ and $(2)$ side by side we have $$\begin{align} DPV \space w(1)-DPV\space w(1)\times\color{red}{\frac{1}{1+r}}=w(1)&+\left[\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+\cdots+\frac{w(1)}{(1+r)^\infty }\right]\\&-\left[\frac{w(1)}{1+r}+\frac{w(1)}{(1+r)^2 }+\cdots+\frac{w(1)}{(1+r)^\infty }\right] \end{align} $$ that is $$\begin{align} DPV w(1)\;\underbrace{\left[1-\frac{1}{1+r}\right]}_{=\frac{r}{1+r}}=w(1) \end{align} $$ and finally $$ DPV w(1)=w(1)\frac{1+r}{r} $$