Fibonacci Fractions

Let

$S = \frac{F_1}{10} + \frac{F_2}{100} + \frac{F_3}{1000} + \cdots = \sum_{n = 1}^{\infty} \frac{F_n}{10^n}$

where $F_n$ denotes the $n$ -th Fibonacci number.

$\begin{aligned} 10S &= F_1 + \frac{F_2}{10} + \frac{F_3}{100} + \cdots \\ 10S - S &= F_1 + \left ( \frac{F_2}{10} - \frac{F_1}{10} \right) + \left ( \frac{F_3}{100} - \frac{F_2}{100} \right ) + \cdots \\ 9S &= 1 + 0 + \frac{F_1}{100} + \frac{F_2}{1000} + \cdots \\ 9S &= 1 + \frac{S}{10} \\ S &= \frac{10}{89}. \end{aligned}$

Thus, $a + b = 10 + 89 = \boxed{99}.$

The generating function for the Fibonacci numbers is

$\frac{x}{1-x-x^{2}} = \sum_{k=1}^{\infty} F_{k}x^{k}$

Plug in $x = \frac{1}{10}$ and you're done.

Oh dear I hated this one but I'm so glad I finished it. First define the $n^{th}$ term as $f(n)$ .

Seeing as this is the sum of Fibonacci numbers, it first makes sense to consider each term as the sum of the previous 2. i.e. $\frac{f(n-2)}{100} + \frac{f(n-1)}{10} = f(n) \implies 100f(n) = 10f(n-1) + f(n-2)$

Let $x = \sum_{n=1}^{\infty}f(n)$ . Now the fun begins by multiplying by 100 $100x = 100\sum_{n=1}^{\infty}f(n) = \sum_{n=1}^{\infty}100f(n) \\ 100x = 11 + \sum_{n=3}^{\infty}100f(n) \\ 100x = 11 + \sum_{n=3}^{\infty}f(n-2) + \sum_{n=3}^{\infty}10f(n-1) \\ 100x = 11 + \sum_{n=1}^{\infty}f(n) + \sum_{n=2}^{\infty}10f(n) \\ 100x = 11 + \sum_{n=1}^{\infty}f(n) + \sum_{n=1}^{\infty}10f(n) - 1 \\ 100x = 11 + x + 10x -1 \\ 89x = 10 \\ x = \frac{10}{89}$

Now that we know what $x$ is, we can now say $a + b = 10 + 89 = \boxed{99}$