Fibonacci is Fun

Solution by algebraic manipulation:

Let $\displaystyle S = \sum_{n=0}^{\infty} \frac{F_{n}}{3^{n}}$ . It can be seen that

$\sum_{n=0}^{\infty} \frac{F_{n}}{3^{n+2}} + \sum_{n=0}^{\infty} \frac{F_{n}}{3^{n+1}} + \frac{F_{0}}{3^{0}} + \frac{F_{1}}{3^{1}} - \frac{F_{0}}{3^{1}} = \sum_{n=2}^{\infty} \frac{F_{n}}{3^{n}}$

since $F_{n-2} + F_{n-1} = F_{n}$ (plus we have some leftover terms), implying

$\frac{S}{9} + \frac{S}{3} + \frac{F_{0}}{3^{0}} + \frac{F_{1}}{3^{1}} - \frac{F_{0}}{3^{1}} = \frac{4S}{9} + \frac{1}{3} = S$

Solving the obtained equation gives us

$S = \sum_{n=0}^{\infty} \frac{F_{n}}{3^{n}} = \boxed{0.6}$

Solution by geometric progression:

Recall that $\displaystyle \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ for all $|x| < 1$ , and that $F_{n} = \frac{\phi^{n}-(-\phi)^{-n}}{\sqrt{5}}$ .

Thus, we can rewrite our sum as

$\sum_{n=0}^{\infty} \frac{F_{n}}{3^{n}} = \frac{1}{\sqrt{5}} \left( \sum_{n=0}^{\infty} \frac{\phi^{n}}{3^{n}} - \sum_{n=0}^{\infty} \frac{(-\phi)^{-n}}{3^{n}} \right)$

Both sums clearly satisfy the criterion $|x| < 1$ , so we can proceed.

$\frac{1}{\sqrt{5}} \left( \sum_{n=0}^{\infty} \frac{\phi^{n}}{3^{n}} - \sum_{n=0}^{\infty} \frac{(-\phi)^{-n}}{3^{n}} \right) = \frac{1}{\sqrt{5}} \left( \frac{1}{1-\frac{\phi}{3}} - \frac{1}{1+\frac{1}{3\phi}} \right)$

After a bit of fiddling, we get that

$\frac{1}{\sqrt{5}} \left( \frac{1}{1-\frac{\phi}{3}} - \frac{1}{1+\frac{1}{3\phi}} \right) = \boxed{0.6}$

who is " Challenge master " ? Is Calvin sir ?