August 2018

Beer 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:

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“

Month: August 2018