Exercise 6.1.7 ($\sum_{j=1}^{k-1} (b_j b_{j+1})^{-1} = 1$)

Question

Consider the fractions $0/1$ to $1/1$ inclusive in the Farey sequence of order $n$. Reading from left to right, let the denominators of these fractions be $b_1,b_2,\ldots,b_k$ so that $b_1 = 1$ and $b_k = 1$. Prove that $\sum_{j=1}^{k-1} (b_j b_{j+1})^{-1} = 1$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $a_j/b_j$ denote the $j$-th fraction. Since $a_j/b_j$ and $a_{j+1}/b_{j+1}$ are adjacent, $b_j a_{j+1} - a_j b_{j+1} = 1$ by Theorem 6.1. Therefore, for all $j \in [\![1, k-1]\!]$,
$$ \frac{a_{j+1}}{b_{j+1}} - \frac{a_j}{b_j} = \frac{1}{b_jb_{j+1}}.$$
Therefore
\begin{align*}
\sum_{j=1}^{k-1}  \frac{1}{b_jb_{j+1}} &= \sum_{j=1}^{k-1} \left(  \frac{a_{j+1}}{b_{j+1}} - \frac{a_j}{b_j}ight)\
&= \sum_{j=2}^{k} \frac{a_j}{b_j} -  \sum_{j=1}^{k-1} \frac{a_j}{b_j}\
&= \frac{a_k}{b_k} - \frac{a_1}{b_1}\
&= 1,
\end{align*}
so
$$\sum_{j=1}^{k-1}  \frac{1}{b_jb_{j+1}} = 1.$$
\end{proof}

Exercise 6.1.7 ($\sum_{j=1}^{k-1} (b_j b_{j+1})^{-1} = 1$)

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Exercise 6.1.7 ($\sum_{j=1}^{k-1} (b_j b_{j+1})^{-1} = 1$)

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