When the source of sound moves away from a stationary listener,we obtain $f_{L} < f_{S}$. Why?

(N/A) The general formula for the Doppler effect is given by $\frac{f_{L}}{v + v_{L}} = \frac{f_{S}}{v + v_{S}}$,where $v$ is the speed of sound,$v_{L}$ is the velocity of the listener,and $v_{S}$ is the velocity of the source.
Given that the listener is stationary,$v_{L} = 0$.
When the source moves away from the listener,the velocity of the source $v_{S}$ is taken as positive in the direction of sound propagation (from listener to source),so the denominator becomes $(v + v_{S})$.
Substituting these values,we get $\frac{f_{L}}{v + 0} = \frac{f_{S}}{v + v_{S}}$.
Therefore,$f_{L} = \left( \frac{v}{v + v_{S}} \right) f_{S}$.
Since $v + v_{S} > v$,the fraction $\frac{v}{v + v_{S}} < 1$. Consequently,$f_{L} < f_{S}$.