Beer theorems and their proofs

TwoBeer.jpgBeer Theorem 1.

An infinite crowd of mathematicians enters a bar.

The first one orders a pint, the second one a half pint, the third one a quarter pint…

“Got it”, says the bartender – and pours two pints.

Proof.

Let s_n = \sum\limits_{i=1}^{n} a \cdot r^{i-1} = a + a\cdot r + a \cdot r^{2} + ...+ a\cdot r^{n-2} + a \cdot r^{n-1}.

Then r\cdot s_{n} = \sum\limits_{i=1}^{n} a\cdot r^{i} = a\cdot r  + a\cdot r^2+ ... + a\cdot r^{n-1} + a\cdot r^{n}

\implies s_{n}-r\cdot s_{n} = a  - a\cdot r^{n}.

Therefore,

s_{n} = {{a\cdot(1-r^{n})} \over {1-r}} .

When a = 1, r={{1} \over {2}},

s_{n} = \frac{1 \cdot (1-({1 \over 2})^n)}{1-{1 \over 2}} = 2\cdot (1-({1 \over 2})^n)

i.e.,

1+ {1 \over 2} + {1 \over 4} + {1 \over 8}+...+({1 \over 2})^{n-1}= 2\cdot (1-({1 \over 2})^n)

\implies \lim\limits_{n \rightarrow \infty} s_{n} = \lim\limits_{n \rightarrow \infty} {2\cdot (1-({1 \over 2})^n)} = 2.

There is also a proof without words at all:

infinite-series-square.jpg

Beer Theorem 2.

An infinite crowd of mathematicians enters a bar.

The first one orders a pint, the second one a half pint, the third one a third of pint…

“Get out here! Are you trying to ruin me?”, bellows the bartender.

Proof.

See “My shot at Harmonic Series

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3 thoughts on “Beer theorems and their proofs

  1. Pingback: My shot at Harmonic Series | Vroom

  2. Pingback: Pandora’s Box | Vroom

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