Can You Get Some Cake?

Riddler Express

You and your infinitely many friends are sharing a cake, and you come up with two rather bizarre ways of splitting it.

For the first method, Friend 1 takes half of the cake, Friend 2 takes a third of what remains, Friend 3 takes a quarter of what remains after Friend 2, Friend 4 takes a fifth of what remains after Friend 3, and so on. After your infinitely many friends take their respective pieces, you get whatever is left.

For the second method, your friends decide to save you a little more of the take. This time around, Friend 1 takes 1/2² (or one-quarter) of the cake, Friend 2 takes 1/3² (or one-ninth) of what remains, Friend 3 takes 1/4² of what remains after Friend 3, and so on. Again, after your infinitely many friends take their respective pieces, you get whatever is left.

Question 1: How much of the cake do you get using the first method?

Question 2: How much of the cake do you get using the second method?

Extra credit: Your friends are feeling rather guilty for not saving enough of the cake for you, so they try one more method. This time, they only take the fractions with even denominators from the second method. So Friend 1 takes 1/2² of the cake, Friend 2 takes 1/4² of what remains, Friend 3 takes 1/6² of what remains after Friend 2, and so on. After your infinitely many friends take their respective pieces, how much of the cake do you get?

Question 1

The $n^{\text{th}}$ friend takes $\dfrac{1}{n+1}$ of the remaining cake. It follows that $\dfrac{n}{n+1}$ of the cake remaining continues to remain after the $n^{\text{th}}$ friend.

Thus, the amount of cake remaining for you is:

$$\prod\limits_{n = 1}^{\infty} \dfrac{n}{n+1} = \dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{3}{4} ... = 0$$

Question 2

Induction will be used to show:

• The fraction of cake remaining for friend $n$ to take from is $\dfrac{n+1}{2n}$
• Friend $n$ takes $\dfrac{1}{2n(n+1)}$ of the whole cake

Friend 1:

• The fraction of cake remaining for Friend $1$ to take from is $\dfrac{1+1}{2 \cdot 1} = 1$ (whole) cake.


• Friend $1$ takes $\dfrac{1}{2 \cdot 1 \cdot (1+1)} = \dfrac{1}{4}$ of the whole cake

Friend $k$:

• Assume the fraction of cake remaining for Friend $k$ to take from is $\dfrac{k+1}{2k}$.


• Friend $k$ takes $\dfrac{1}{(k+1)^2}$ of that. Friend $k$ takes $\dfrac{k+1}{2k} \cdot \dfrac{1}{(k+1)^2} = \dfrac{1}{2k(k+1)}$ of the whole cake.

Friend $k+1$:

• Since we know what Friend $k$ took, the fraction of cake remaining for Friend $k+1$ is:
  
  $$\dfrac{k+1}{2k} \cdot \dfrac{(k+1)^2-1}{(k+1)^2}$$
  $$ = \dfrac{k+1}{2k} \cdot \dfrac{k(k+2)}{(k+1)(k+1)}$$
  $$ = \dfrac{k+2}{2(k+1)}$$


• Friend $k+1$ takes $\dfrac{k+2}{2(k+1)} \cdot \dfrac{1}{(k+2)^2} = \dfrac{1}{2(k+1)(k+2)}$ of the whole cake.

Since it has been proven that the amount of cake remaining is $\dfrac{n+1}{2n}$, you get:

$$\lim_{n \to \infty} \dfrac{n+1}{2n} = \dfrac{1}{2} \text{ of the cake.}$$

Extra Credit

The $n^{\text{th}}$ friend takes $\dfrac{1}{4n^2}$ of the cake remaining. It follows that $\dfrac{4n^2-1}{4n^2}$ of the cake remaining continues to remain after the $n^{\text{th}}$ friend.

Thus, the amount of cake remaining for you is:

$$\prod\limits_{n = 1}^{\infty} \left( \dfrac{4n^2 - 1}{4n^2} \right)$$

$$ = \dfrac{1 \cdot 3}{2 \cdot 2} \cdot \dfrac{3 \cdot 5}{4 \cdot 4} \cdot \dfrac{5 \cdot 7}{6 \cdot 6} \cdot \dfrac{...}{...}$$

This is the reciprocal of the Wallis product, so you get $\dfrac{2}{\pi}$ of the cake.

Rohan Lewis

2021.05.17