$F_k=F_{k1}+F_{k2}$
where $F_0=0, F_1=1$
. Define the series $s(x)= \sum_{k=1}^\infty F_k / x^k$
for $x>2$
. Show that $s(x)=\frac{x}{x^2x1}$
.

Use the recursive definition of
$F_k$
to split$s(x)$
into two sums. Carefully treat the base cases. 
Try rewriting the summation in terms of
$s(x).$

You may relabel the index of each sum in order to get it to match the definition of
$s(x).$

The key is to breakdown
$F_k$
using its recursive definition, then relabelling the indices to get$s(x)$
again. The first term of the sum$\frac{F_1}{x}$
cannot be broken down as it is a base case, and must be taken out of the sum first.$$ \begin{aligned} s(x)&=\sum_{k=1}^\infty \frac{F_k}{x^k}\\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_k}{x^k} \\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_{k1}}{x^k} + \sum_{k=2}^\infty \frac{F_{k2}}{x^k} \\ &=\frac{F_1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{F_0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \frac{1}{x}\sum_{i=1}^\infty \frac{F_{i}}{x^i} + \frac{1}{x^2} \sum_{j=1}^\infty \frac{F_{j}}{x^j} \\ &=\frac{1}{x} + \frac{1}{x}s(x) + \frac{1}{x^2}s(x) \end{aligned} $$
Finally, we rearrange to get
$s(x)=\frac{x}{x^2x1}$
.