Fractal Fractions

let,

$\frac { 3 }{ 4 } +\frac { \frac { 3 }{ 4 } +\frac { \frac { 3 }{ 4 } +\frac { \frac { 3 }{ 4 } +\frac { ... }{ 4 } }{ 4 } }{ 4 } }{ 4 } =x$

Since the fraction goes on for infinity,

$x=\frac { 3 }{ 4 } +\frac { x }{ 4 } =\frac { 3+x }{ 4 }$

Solving it gives $x=1$

Therefore, the answer is 1

Consider a sequence of ${a_n}$ , where:

$a_n = \frac{3+a_{n-1}}{4}$

$\lim_{n\to\infty} a_n = L$ because we can assume this fraction converges to a limit (otherwise we wouldn't be able to do this problem), we know that as $n\to\infty, a_n = a_{n+1}$

So:

$L = \lim_{n\to\infty} a_{n+1}\\$ apply the recursive definition: $L = \lim_{n\to\infty} \frac{3+a_n}{4}\\ L = \frac{3+L}{4}\\ 4L = 3+L\\ 3L = 3\\ L = 1$ $\lim_{n\to\infty} a_n = 1$