Continued Fraction Expansion of Irrational Square Root/Examples/61/Convergents

Convergents to Continued Fraction Expansion of $\sqrt {61}$
The sequence of convergents to the continued fraction expansion of the square root of $61$ begins:
 * $\dfrac 7 1, \dfrac {8} 1, \dfrac {39} 5, \dfrac {125} {16}, \dfrac {164} {21}, \dfrac {453} {58}, \dfrac {1070} {137}, \dfrac {1523} {195}, \dfrac {5639} {722}, \dfrac {24079} {3083}, \ldots$

Proof
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be its continued fraction expansion.

Let $(p_n)_{n\geq 0}$ and $(q_n)_{n\geq 0}$ be its numerators and denominators.

Then the $n$th convergent is $p_n/q_n$.

By definition:


 * $p_k = \begin{cases} a_0 & : k = 0 \\

a_0 a_1 + 1 & : k = 1 \\ a_k p_{k - 1} + p_{k - 2} & : k > 1\end{cases}$


 * $q_k = \begin{cases} 1 & : k = 0 \\

a_1 & : k = 1 \\ a_k q_{k - 1} + q_{k - 2} & : k > 1\end{cases}$

From Continued Fraction Expansion of $\sqrt {61}$:
 * $\sqrt {61} = \left[{7, \left\langle{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}\right\rangle}\right]$

Thus the convergents are assembled:


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