Three students share a bread. The first eats half, the second eats half of what is left, and the third eats half of that and so on....How much does each eat?
student 1 = $\displaystyle \frac{1}{2}+\frac{1}{16}+\frac{1}{64}+...=\frac{1} {2^1}+\frac{1}{2^4}+\frac{1}{2^7}+...=\sum_{n=0}^{ \infty}\frac{1}{2^{3n+1}}=$
$\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{8^n}=\frac{ 1}{2}\left( \frac{1}{1-\frac{1}{8}}\right)=\frac{1}{2} \left( \frac{1}{\frac{7}{8}}\right)=\frac{1}{2} \cdot \frac{8}{7}=\frac{4}{7}$
You will find a similar pattern for students 2 and 3.
The fractions consumed by each
1 are 1/2 , 1/16 1/128 totalling ˝[1+(1/8) + (1/8)2 +…]
2 are 1/4 , 1/32, 1/256 totalling 1/4[1+(1/8) + (1/8)2 +…]
3 are 1/8, 1/64 1/512, totalling 1/8[1+(1/8) + (1/8)2 +…]
[1+(1/8) + (1/8)2 +…] = 1 / (1-1/8) = (8/7). Therefore,
1 consumes (1/2) (8/7) = 4/7
2 consumes (1/4) (8/7) = 2/7
3 consumes (1/8) (8/7) = 1/7
Note 4/7 + 2/7 + 1/7 = 1