Fibonacci sequence and other metallic sequences emerged in the form of fractions

Answer to question 1) :

The generating function for the Fibonacci numbers $F_n$ is known to be

$$\dfrac{1}{1-(x+x^2)}=\underbrace{1}_{F_0}+\underbrace{1}_{F_1}x+\underbrace{2}_{F_2}x^2+\underbrace{3}_{F_3}x^3+\underbrace{5}_{F_4}x^4+\cdots+F_nx^n+...$$

Taking $x=0.1$ gives :

$$\dfrac{1}{1-0.11}=1+1 \times 0.1+2 \times 0.01+3 \times 0.001+5 \times 0.0001+\cdots+F_n 0.1^n+...$$

justifying the equality of LHS and RHS of your first identity (multiplied by $100$).

Same process for the other metallic sequences.

For example, the generating functions of the silver and bronze sequences are resp.

$$\dfrac{1}{1-(2x+x^2)} \ \ \ \text{and} \ \ \ \dfrac{1}{1-(3x+x^2)}$$

An interesting generalization along these lines : the recent paper https://arxiv.org/pdf/1901.02619.pdf

Following on from Jean Marie's answer, the metallic sequence $$M_{n,k}=nM_{n,k-1}+M_{n,k-2}$$ Has the generating function $$G_n(x)=M_{n,0}+M_{n,1}x+M_{n,2}x^2+\dots$$ Such that $$xG_n(x)=M_{n,0}x+M_{n,1}x^2+M_{n,2}x^3+\dots$$ $$nG_n(x)=nM_{n,0}+nM_{n,1}x+nM_{n,2}x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+(nM_{n,1}+M_{n,0})x+(nM_{n,2}+M_{n,1})x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+M_{n,2}x+M_{n,3}x^2+\dots$$ $$(x+n)G_n(x)=nM_{n,0}+\frac{G_n(x)-M_{n,0}}x-M_{n,1}$$ $$x(x+n)G_n(x)=nM_{n,0}x+G_n(x)-M_{n,0}-M_{n,1}x$$ $$(x(x+n)-1)G_n(x)=(nM_{n,0}-M_{n,1})x-M_{n,0}$$ $$G_n(x)=\frac{M_{n,0}+(M_{n,1}-nM_{n,0})x}{1-x(x+n)}$$ But we have the values of $M_{n,0}=0$ and $M_{n,1}=1$ hence this just becomes $$G_n(x)=\frac{x}{1-x(x+n)}$$ If we let $x=\frac1{10}$ we get the fractional representation as mentioned, $$G_n\left(\frac1{10}\right)=\frac{1/10}{1-(1/10+n)/10}=\frac{10}{99-10n}$$ Which gives fractional values of $$\frac{10}{89},\frac{10}{79},\frac{10}{69},\frac{10}{59},\dots$$ and each of these still contain the corresponding metallic sequence as, for example, $$\frac{10}{89}=0.\overline{11235955056179775280898876404494382022471910}$$