Fibonacci on fives

A Solution Using The generating function of the Fibonacci Sequence :

With X = 1 / 5

We Get S = 5 / 19

Then A = 5 & B = 19

Finally A + B = 24

$S =\frac{1}{5} + \frac{1}{5^{2}} + \frac{2}{5^{3}} + \frac{3}{5^{4}} + \frac{5}{5^{5}} + \frac{8}{5^{6}} + \cdots$ $5S =1+\frac{1}{5} + \frac{2}{5^{2}} +\frac{3}{5^{3}} +\frac{5}{5^{4}} + \cdots$ $4S=1+\frac{1}{5^{2}} +\frac{1}{5^{3}} +\frac{2}{5^{4}} +\frac{3}{5^{5}} +\cdots$ $4S= 1+\frac{1}{5} \left( \frac{1}{5} +\frac{1}{5^{2}} +\frac{2}{5^{3}} +\frac{3}{5^{4}} +\frac{5}{5^{5}} +\cdots \right)$ $4S=1+\frac{S}{5}$ So $S=\frac {5}{19}$ Now $A+B = \boxed {24}$

$S =\frac{1}{5} + \frac{1}{5^{2}} + \frac{2}{5^{3}} + \frac{3} {5^{4}} + \frac{5}{5^{5}} + \ldots \infty \\ \frac{S}{5}=\frac{1}{5^{2}} + \frac{1}{5^{3}} + \frac{2}{5^{4}} + \frac{3} {5^{5}} + \frac{5}{5^{6}} + \ldots \infty \\ \frac{4S}{5}=\frac{1}{5}+\underbrace{\frac{1}{5^{3}}+\frac{1}{5^{4}}+\frac{2}{5^{5}}+\frac{3}{5^{6}}+\ldots \infty}_{\frac{S}{5^{2}}} \\ \Rightarrow \frac{4S}{5}=\frac{1}{5}+\frac{S}{5^{2}}\Rightarrow \frac{19S}{25}=\frac{1}{5}\Rightarrow S=\frac{5}{19} \\ \Rightarrow 5+19=\boxed{24}$

Let $S = \frac{1}{5}+ \frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...$ We note that denominators holds the following recurrence $a_n+a_{n+1}=a_{n+2}; a_1 = a_2 = 1$ that is : $S = \frac{2-1}{5}+ \frac{3-2}{5^2}+\frac{5-3}{5^3}+\frac{8-5}{5^4}+...$ Spliting from positive to negative terms and rearranging the sum : $S = (\frac{2}{5}+ \frac{3}{5^2}+\frac{5}{5^3}+\frac{8}{5^4}+... )$ - $(\frac{1}{5}+ \frac{2}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+... )$ We observe that : $S = (25S-6) - (5S-1)$ Finally : $S = \frac{5}{19}$

But not too fast! ... First we need to decide if this sums converges, because calculating a value, really doesn't mean that series converges to that value and even doesn't mean that we can rearrange terms . We use ratio test and convergence ratio of fibonacci secuence for this purpose :

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n} \frac{1}{5}| = \frac{ \varphi}{5} < 1$ Therefore is absolutely convergent.