Infinite Fractions - 6

$\large \cfrac { 8 }{ \sqrt {\color{#3D99F6}{ \cfrac { 8 }{ \sqrt { \cfrac { 8 }{ \sqrt { \cfrac { 8 }{ \sqrt { \cfrac { 8 }{ \sqrt { \ddots } } } } } } } } } } } = \,\color{#3D99F6}{x}$ $\large \cfrac { 8 }{\sqrt{\color{#3D99F6}{x}}} = \color{#3D99F6}{x}$ $\large 8=\sqrt{x^3}$ $\large 64={x^3}$ $\large 4=x$

The exponent is the value of a geometric series: $8^{1 - 1/2 + 1/4 - 1/8 + \cdots} =8^{2/3} = 4$ .

The nested radical function in question is equal to

$8 \div (8^{1/2} \div (8^{1/2 \times 1/2} \div (8^{1/2 \times1/2 \times1/2} \div (8^{1/2 \times1/2 \times1/2\times12} \div ( \cdots ))) \cdots )$

For consistency, we convert all the divisions to multiplications to get:

$\large 8^1 \times 8^{-1/2} \times 8^{1/2 \times1/2} \times 8^{-1/2 \times 1/2 \times 1/2} \times \cdots = 8^{1 - \frac12 + \frac14 - \frac18 + \cdots}$

This is because $\large a^{p_1} \times a^{p_2} \times a^{p_3} \times \cdots = a^{p_1 + p_2 + p_3 + \cdots}$ .

What's left to do is to evaluate the infinite geometric progression sum , $1 - \frac12 + \frac14 - \frac18 + \cdots$ . In this case, the first term $a = 1$ , common ratio $r = -\dfrac12$ , and thus the sum is $\dfrac a{1-r} =\dfrac1{1 + \frac12} =\dfrac23$ .

So the nested function in question is equal to $8^{2/3} = (2^3)^{2/3} = 2^{3\times 2/3} = 2^2 = \boxed4$ .