Question about the algorithm. A chocolate company decides to promote its chocolate bars by including a coupon
with each bar. A bar costs a dollar and with c coupons you get a new bar.
How much chocolate is a dollar worth?
As from my understanding:
$1$ Bar $= \$1$ , $C$ coupons $=  1$ bar
$1$ Bar $= \$1 - C$ coupons , when it says how much chocolate is a dollar worth.
Is it $\$1= 1$ Bar $+ C$ coupons?
Can someone help me out?
 A: We do the analysis in two ways, the first a little tedious, the second not. 
Tedious: For our dollar, we get $1$ bar and $1$ coupon. The coupon is worth 
$\frac{1}{c}$ 
of a chocolate bar, and $\frac{1}{c}$ of a coupon. The $\frac{1}{c}$ of a coupon is worth 
$\frac{1}{c^2}$
of a chocolate bar, and $\frac{1}{c^2}$ of a coupon. That coupon is worth
$\frac{1}{c^3}$
of a chocolate bar and $\dots$.
Thus, if our chocolate eater lives forever, eating enormous quantities of chocolate, and the coupon policy goes on forever, the $\$1$ will ultimately buy
$$1+\frac{1}{c}+\frac{1}{c^2}+\frac{1}{c^3}+\cdots$$
bars of chocolate, Using the ordinary formula for the sum of an infinite geometric series, we get that this is $\frac{c}{c-1}$.
Not tedious:  Let the amount of chocolate a dollar is worth be $v$. Then $1$ dollar gets us a chocolate bar, plus a coupon worth $\frac{v}{c}$. Thus
$$v=1+\frac{v}{c}.$$
Solve this linear equation for $v$. We get $v\left(1-\frac{1}{c}\right)=1$, which after some algebra yields $v=\frac{c}{c-1}$. 
