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The numbers $k_{n}$ and $k_{n+1}$ , as $n$ gets larger ( $n \rightarrow \infty$ ), should have less and less difference between them, ie $k_{n} \sim k_{n+1} \sim \frac{1+k_{n}}{4}$ .

Alternatively, think of $k_{n}$ as

$k_{n} = \frac{1+\frac{1+\frac{1+\ldots}{4}}{4}}{4} = \frac{1+k_{n}}{4}$

By these two perspectives, solving for $k_{n}$ , we get

$k_{n} = \frac{1+k_{n}}{4} \longrightarrow k_{n} = \boxed{\frac{1}{3}}$

A rigorous proof :

$k_{2}=\frac{1}{4}k_1+\frac{1}{4}$

$k_{3}=\frac{1}{4}k_2+\frac{1}{4}=\frac{1}{4^2}k_1+\frac{1}{4}+\frac{1}{4^2}$

$k_{n+1}=\frac{1}{4^n}k_1+\sum_{i=1}^{n}\frac{1}{4^i}$

Thus,

$\lim_{n \to \infty} k_{n+1} = \lim_{n \to \infty}\frac{1}{4^n}k_1+\sum_{i=1}^{n}\frac{1}{4^i} =\frac{1/4}{1-1/4}=\frac{1}{3}=\frac{a}{b}$

Thus, $a+b=1+3=\boxed{4}$

Trying the problem from a mortal's perspective helps approach the right answer:

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from fractions import Fraction as frac
board = []
terms = 100
for i in xrange(terms):
    board.append(1)
def new(b):
    b[0] = frac(b[0]+b[1],4)
    del b[1]
    return b
while len(board) > 1:
    board = new(board)
print "Answer:", float(board[0])

Output:

Answer: 0.333333333333