Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^\frac{1}{2}}^k}{2}$

Question

While doing questions on series and products, I tried hard to solve this infinite product, but no methods has worked so far.Well, this is from Johns Hopkins math tournament

Question:- Evaluate $\prod_{k=1}^{\infty}\frac{{5^\frac {1}{2}}^k+{3^\frac{1}{2}}^k}{2}$

Can anybody help me out!

Answer 1

Let $\alpha = 5, \beta = 3$. Recall the identity

$$x + y = \frac{x^2 - y^2}{x-y}$$ The product you have is a telescoping one.

$$\prod_{k=1}^N \frac{\alpha^{2^{-k}} + \beta^{2^{-k}}}{2} = \frac{1}{2^N} \prod_{k=1}^N \frac{\alpha^{2^{1-k}} - \beta^{2^{1-k}}}{\alpha^{2^{-k}} - \beta^{2^{-k}}} = \frac{\alpha - \beta}{\frac{\alpha^{2^{-N}} - \beta^{2^{-N}}}{2^{-N}}} \tag{*1} $$ Since $2^{-N} \to 0$ as $N \to \infty$, we can use the fact

$$\lim_{x\to0} \frac{u^x - 1}{x} = \left.\frac{du^x}{dx}\right|_{x=0} = \log u\quad\text{ for }\quad u = \alpha,\beta $$ to compute the limit of the denominator in $(*1)$. $$\lim_{N\to\infty} \frac{\alpha^{2^{-N}} - \beta^{2^{-N}}}{2^{-N}} = \lim_{x\to 0} \frac{\alpha^x - \beta^x}{x} = \lim_{x\to 0} \left[\frac{\alpha^x - 1}{x} - \frac{\beta^x - 1}{x}\right] = \log\frac{\alpha}{\beta}$$ As a result,

$$\prod_{k=1}^\infty \frac{5^{2^{-k}} + 3^{2^{-k}}}{2} = \frac{5 - 3}{\log\frac53} = \frac{2}{\log\frac{5}{3}} \approx 3.91523037794...$$