ProofWiki:Sandbox

Proof
From the recursive definition of continued fractions, we have:

Let:

In other words, $a_{3n+1} = 2n$ and $a_{3n+0} = a_{3n+2} = 1$

Then $p_i$ and $q_i$ are as follows:


 * $\begin{array}{r|cccccccccc}

\displaystyle i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9  \\ \hline

p_i & 1 & 1 & 2 & 3 & 8 & 11 & 19 & 87 & 106 & 193 \\

q_i & 1 & 0 & 1 & 1 & 3 & 4 & 7 & 32 & 39 & 71 \\ \hline

\end{array}$

Furthermore, $p_i$ and $q_i$ satisfy the following $6$ recurrence relations:

Our ultimate aim is to prove that:

In the pursuit of that aim, let us define the integrals:

Lemma 1 - separate page


 * We assert for $n \in \Z, n \ge 0$,

Include up to here

To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:

The final step needed to validate the assertion, we must demonstrate that the following three recurrence relations hold:

To prove the first relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n-1}$ and the integrand of $C_{n-1}$

By integrating both sides of the equation, we verify the first recurrence relation:

To prove the second relation, we note that the derivative of the integrand of $C_n$ is equal to the sum of the integrand of $B_n$ with two times n times the integrand of $A_{n}$ minus the integrand of $C_{n-1}$

By integrating both sides of the equation, we verify the second recurrence relation:

To prove the third relation, we have:

From the first relation, combined with the initial condition at $n = 0$ being satisfied, we have:

From the second relation, combined with the initial condition at $n = 0$ being satisfied, we have:

From the third relation, combined with the initial condition at $n = 0$ being satisfied, we have:

We also assert that $A_n$, $B_n$ and $C_n$ all converge to $0$ as $n \mathop \to \infty$:

We now have:

From which, we conclude: