Let $F_k=F_{k-1}+F_{k-2}$ 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 $s(x)=\frac{x}{x^2-x-1}$.

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. $$ \begin{aligned} s(x)&=\sum_{k=1}^\infty \frac{F_k}{x^k}\\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_k}{x^k} \\ &=\frac{F_1}{x} + \sum_{k=2}^\infty \frac{F_{k-1}}{x^k} + \sum_{k=2}^\infty \frac{F_{k-2}}{x^k} \\ &=\frac{F_1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{F_0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \sum_{i=1}^\infty \frac{F_{i}}{x^{i+1}} + \bigg(\frac{0}{x^2} + \sum_{j=1}^\infty \frac{F_{j}}{x^{j+2}}\bigg) \\ &=\frac{1}{x} + \frac{1}{x}\sum_{i=1}^\infty \frac{F_{i}}{x^i} + \frac{1}{x^2} \sum_{j=1}^\infty \frac{F_{j}}{x^j} \\ &=\frac{1}{x} + \frac{1}{x}s(x) + \frac{1}{x^2}s(x) \end{aligned} $$

Finally, we rearrange to get $s(x)=\frac{x}{x^2-x-1}$.