Telescoping product

For $y > 0$ , define the sequence $G_n$ by the recurrence relation $G_0 \; = \; 0\;,\; G_1 \; = \; 1 \;, \; \hspace{2cm} G_{n+1} \; = \; yG_n + G_{n-1} \qquad (n \ge 1)$ Then it is a simple induction that the sequence of numerators $x_n$ given by the recurrence relation $x_1 \; = \; y \;,\; \hspace{2cm} x_{n+1} \; = \; y + x_n^{-1} \qquad (n \ge 1)$ is given by the formula $x_n \; = \; \frac{G_{n+1}}{G_n} \hspace{2cm} n \ge 1$ We can obtain a generalisation of Binet's formula, in that $G_n \; = \; \frac{1}{\sqrt{y^2+4}} \left[\left(\frac{y + \sqrt{y^2+4}}{2}\right)^n - \left(\frac{y - \sqrt{y^2+4}}{2}\right)^n\right] \; = \; \frac{\phi(y)^n - (-\phi(y)^{-1})^n}{\sqrt{y^2+4}} \hspace{2cm} n \ge 0$ where $\phi(y) \; = \; \frac{y + \sqrt{y^2+4}}{2} > 1$ From this it is clear that $A(y) \; = \; \lim_{n \to \infty} x_n \; = \; \lim_{n \to \infty}\frac{G_{n+1}}{G_n} \; =\; \phi(y)$ and that $\prod_{n=1}^\infty \frac{x_n}{A(y)} \;= \; \lim_{N \to \infty}\prod_{n=1}^N \frac{x_n}{A(y)} \; = \; \lim_{N \to \infty}\frac{G_{N+1}}{G_1 \phi(y)^N} \; = \; \frac{\phi(y)}{\sqrt{y^2+4}} \; = \; \frac{y + \sqrt{y^2+4}}{2\sqrt{y^2+4}}$ so that, in general, $a=y$ , $b = y^2+4$ and $c=2$ . In the case $y=1$ , this makes the answer $\boxed{152}$ .

if $y < 0$ , all the $x_n$ terms change sign, and hence the limit $A$ also changes sign (so that $A(-y) = -A(y)$ ). This means that the sequence of finite products converges to $\frac{|y| + \sqrt{y^2+4}}{2\sqrt{y^2+4}}$ for all nonzero $y$ .

Define the truncated version of $A(x)$ as

$A_k(x)=\underbrace{x+\dfrac{1}{x+\dfrac{\ddots}{x+\frac{1}{x}}}}_{k\text{ x's}}$

The product we have to find is

$P(1)=\displaystyle\prod_{k=1}^{\infty}\dfrac{A_k(1)}{A(1)}$

Claim:

$A_k(1)=\dfrac{F_{k+1}}{F_k}$

Where $F_k$ is the $k$ -th Fibonacci number

Proof:

Base case: $k=1$

$A_k(1)=1=\dfrac{1}{1}=\dfrac{F_2}{F_1}$

Induction step: $k\Longrightarrow k+1$

Since

$A_{k+1}(1)=1+\dfrac{1}{A_k(1)}$

It follows that

$\begin{aligned} A_{k+1}(1)&=1+\dfrac{1}{\frac{F_{k+1}}{F_k}}\\ &=1+\dfrac{F_k}{F_{k+1}}\\ &=\dfrac{F_{k+1}+F_k}{F_{k+1}}\\ &=\dfrac{F_{k+2}}{F_{k+1}}\phantom{cccc}\square \end{aligned}$

Which also gives the value of $A(1)$

$A(1)=\displaystyle\lim_{k\rightarrow\infty}A_k(1)=\displaystyle\lim_{k\rightarrow\infty}\dfrac{F_{k+1}}{F_k}=\phi$

Considering a finite version of the product

$P_n(1)=\displaystyle\prod_{k=1}^n \dfrac{A_k(1)}{\phi}=\dfrac{1}{\phi^n}\left(\dfrac{F_2}{F_1}\cdot\dfrac{F_3}{F_2}\cdot\cdot\cdot\dfrac{F_{n+1}}{F_n}\right)=\dfrac{F_{n+1}}{\phi^n}$

Hence

$\begin{aligned} P(1)&=\displaystyle\lim_{n\rightarrow\infty}P_n(1)\\ &=\displaystyle\lim_{n\rightarrow\infty}\dfrac{F_{n+1}}{\phi^n}\\ &=\displaystyle\lim_{n\rightarrow\infty}\dfrac{\frac{1}{\sqrt{5}}(\phi^{n+1}-(-\phi)^{-(n+1)})}{\phi^n}\phantom{cccccccccc}\color{#3D99F6}\text{Binet's Formula}\\ &=\dfrac{\phi}{\sqrt{5}}=\dfrac{1+\sqrt{5}}{2\sqrt{5}}=\dfrac{a+\sqrt{b}}{c\sqrt{b}} \end{aligned}$

$\Longrightarrow 100a+10b+c=\boxed{152}$

First, let's try to write a closed form for $A(x)$ . First we notice $A(x)=x+\frac{1}{A(x)} \\ (A(x))^2=xA(x)+1 \\ (A(x))^2-xA(x)-1=0$ By the quadratic formula, we get $A(x)=\frac{x \pm \sqrt{x^2+4}}{2}$ Since in the original equation, we are adding a number to $1$ , we can conclude that the $\pm$ should be $+$ . Using our new closed form of $A(x)$ , we get that $A(1)=\frac{1+\sqrt{1^2+4}}{2}=\frac{1+\sqrt{5}}{2}$ Now, looking at the infinite product, we notice that the reciprocal of $A(1)$ is multiplied an infinite number of times, so we can take it out and try to evaluate the numerator. Looking at the first few terms, we notice, whenever we go up to the other term, we are simply taking the reciprocal of the previous term and adding $1$ . In other words $a_n=\frac{1}{a_{n-1}}+1 = \frac{a_{n-1}+1}{a_{n-1}}$ Since the terms can always be written in fractional form, we can look at the numerator and denominator separately, so $\frac{a_{n-1}+1}{a_{n-1}}=\frac{\frac{p_{n-1}}{q_{n-1}}+\frac{q_{n-1}}{q_{n-1}}}{\frac{p_{n-1}}{q_{n-1}}}=\frac{p_{n-1}+q_{n-1}}{p_{n-1}} \textrm{ where p is the numerator and q is the denominator}\\$ Notice, the numerator now forms the Fibonacci Sequence. Also, notice, the numerator of the first term and denominator of the second term cancel out when we multiply them, so this product is simply the infinith Fibonacci number. We can evaluate this as a limit, namely $\lim_{n \to \infty} \frac{1}{\sqrt{5}}( \frac{1+\sqrt{5}}{2})^{n+1} \textrm{ Which is an approximation of the nth Fibonacci Number}$ Now putting this into the overall product, we get $\lim_{n \to \infty} (\frac{2}{1+\sqrt{5}})^n \cdot \frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^{n+1}=\frac{1}{\sqrt{5}} \cdot (\frac{1+\sqrt{5}}{2})=\frac{1+\sqrt{5}}{2\sqrt{5}} \implies a=1, b=5, c=2, \textrm{ so } 100a+10b+c=100+50+2=\boxed{152}$

Bonus: From what we did above, we can simply get that the closed formula for $\frac{y}{A(y)} \times \frac{y+\frac{1}{y}}{A(y)} \times \dots = \lim_{n \to \infty} (\frac{1}{A(y)})^n \times \frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^{n+1}=\boxed{\lim_{n \to \infty}(\frac{1+\sqrt{5}}{2\sqrt{5}}) \cdot (\frac{1+\sqrt{5}}{|y|+\sqrt{y^2+4}})^n}$