Farey Sequence is not Convergent

Theorem

Consider the Farey sequence:

$F = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$

$F$ is not convergent.

Proof

We have the following subsequences of $F$ which are all convergent to a different limit:

$\dfrac 1 2, \dfrac 2 4, \dfrac 3 6, \dfrac 4 8 \to \dfrac 1 2$ as $n \to \infty$

$\dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \dfrac 1 5 \to 0$ as $n \to \infty$

$\dfrac 1 2, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5 \to 1$ as $n \to \infty$

It follows from Limit of Subsequence equals Limit of Real Sequence that $F$ cannot be convergent.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Bolzano-Weierstrass Theorem: $\S 5.12$: Example $\text{(ii)}$