1. another fibonacci infinite sum

Soroban, here's another fun fibonacci sum for ya'.

Prove that:

$\frac{1}{F_{1}}+\frac{1}{F_{2}}+\frac{1}{F_{4}}+\f rac{1}{F_{8}}+\frac{1}{F_{16}}+............... = 4-{\tau}$

In other words, prove: $\sum_{n=0}^{\infty}\frac{1}{F_{2^{n}}}=4-{\tau}$

As you probably know, $\tau$ is the golden ratio.

golden ratio = $\frac{1+\sqrt{5}}{2}$

2. Rather anyone cares or not, but here is how I tackled this.

I started with:

$\frac{F_{2^{n}+1}+2F_{2^{n}-2}}{F_{2^{n}}}=\frac{F_{2^{n}+1}}{F_{2^{n}}}+\frac {2F_{2^{n}-2}}{F_{2^{n}}}$

Now, $\lim_{n\to\infty}\frac{F_{2^{n}+1}}{F_{2^{n}}}={\t au}$

So it's left to show that $\lim_{n\to\infty}\frac{2F_{2^{n}-2}}{F^{2^{n}}}=3-\sqrt{5}$.

Because ${\tau}+3-\sqrt{5}=4-{\tau}$

Based on the observation that $\frac{{\tau}^{n}-{\mu}^{n}}{{\tau}-{\mu}}$ = the Fibonacci sequence, and ${\mu}=1-{\tau}$

Let $k=2^{n}$, then $\frac{2F_{k-2}}{F_{k}}$

$F_{k}=\frac{{\tau}^{k}-{\mu}^{k}}{{\tau}-{\mu}}$

and $F_{k-2}=\frac{{\tau}^{k-2}-{\mu}^{k-2}}{{\tau}-{\mu}}$

Then we get: $\frac{2({\tau}^{k-2}-{\mu}^{k-2})}{{\tau}^{k}-{\mu}^{k}}$

But $\lim_{k\to\infty}{\mu}^{k}=0$

So, we are left with $\frac{2{\tau}^{k-2}}{{\tau}^{k}}=\frac{2}{{\tau}^{2}}=3-\sqrt{5}$

And we have $\sum_{n=1}^{\infty}\frac{1}{F_{2^{n}}}=4-{\tau}$