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