One Third as Quotient of Sequences of Odd Numbers

Theorem

 * $\dfrac 1 3 = \dfrac {1 + 3} {5 + 7} = \dfrac {1 + 3 + 5} {7 + 9 + 11} = \dfrac {1 + 3 + 5 + 7} {9 + 11 + 13 + 15} = \cdots$

That is:
 * $\forall n \in \Z_{> 0}: \dfrac 1 3 = \dfrac {\ds \sum_{k \mathop = 1}^n \paren {2 k - 1} } {\ds \sum_{k \mathop = n + 1}^{2 n} \paren {2 k - 1} }$

Proof
The result follows.