Joy of Mathematics

A finite product

April 10, 2023 Tolaso

Evaluate the product

$\Pi = \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1}$

Solution

We are making use of the identity

$x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) = (x^2 + x + 1)((x-1)^2 + (x-1) + 1)$

Hence,

$\begin{align*} \prod_{k=1}^{n}\frac{(2k)^4+(2k)^2+1}{(2k-1)^4+(2k-1)^2+1} &= \prod_{k=1}^{n} \frac{\cancel{\left ( 4k^2- 2k+1 \right )} \left ( 4k^2+2k+1 \right )}{\left ( 4k^2-6k+3 \right ) \cancel{\left ( 4k^2-2k+1 \right )}} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=2}^{n} \left ( 4k^2-6k+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4(k+1)^2-6(k+1)+3 \right )} \\ &= \frac{\prod_{k=1}^{n} \left ( 4k^2 + 2k+1 \right )}{\prod_{k=1}^{n-1} \left ( 4k^2 + 2k+1 \right )} \\ &= 4n^2 +2n+1 \end{align*}$

An eta Dedekind type product

March 16, 2023 Tolaso

Prove that

$\prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) = \frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}}$

Solution

Let $\tau = \frac{i}{\sqrt{3}}$ and $q = \exp \left ( 2 \pi i \tau \right )$ . Hence,

$\begin{align*} \prod_{n=1}^{\infty} \left (1 + 2e^{-n \pi \sqrt{3}} \cosh \frac{n \pi}{\sqrt{3}} \right ) &= \prod_{n=1}^{\infty} \left ( 1 + q + q^{2n} \right ) \\ &= \frac{\prod_{n=1}^{\infty} \left ( 1 - q^{3n} \right )}{\prod_{n=1}^{\infty} \left ( 1 - q^n \right )} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta(3\tau)}{\eta(\tau)} \\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \frac{\eta \left ( - 1/\tau \right )}{\eta(\tau)}\\ &= \exp\left ( - \frac{i \pi \tau}{6}\right ) \sqrt{-i \tau} \\ &= frac{\exp \left ( \frac{\pi \sqrt{3}}{18} \right )}{\sqrt[4]{3}} \end{align*}$

An infinite product with Lucas and Fibonacci numbers

March 4, 2023March 4, 2023 Tolaso

Let $\mathcal{F}_n$ denote the $n$ -th Fibonacci number and $\mathcal{L}_n$ the $n$ – th Lucas. Prove that

$\prod_{n=1}^{\infty} \left ( 1 + \frac{1}{\mathcal{F}_{2^n +1} \mathcal{L}_{2^n+1}} \right ) = \frac{3}{\varphi^2}$

Solution

Exploiting the fact that $\mathcal{F}_n \mathcal{L}_n = \mathcal{F}_{2n}$ we deduce that $\mathcal{F}_{2^n+1} \mathcal{L}_{2^n+1}=\mathcal{F}_{2^{n+1}+2}$ .

Lemma: Let $a, b$ be positive numbers. It holds that

$\prod_{n=1}^{\infty} \left ( 1 + \frac{\mathcal{F}_b}{\mathcal{F}_{2^n a + b}} \right ) = \frac{1 - (-1)^b \varphi^{-2a-2b}}{1 - \varphi^{-2a}}$

Proof: Making use of Binet’s formula we have successively:

$\begin{align*} \prod_{n=1}^{\infty} \left ( 1 + \frac{\mathcal{F}_b}{\mathcal{F}_{2^n a + b}} \right ) &= \prod_{n=1}^{\infty} \left ( 1 + \frac{\varphi^b - (-1)^b \varphi^{-b}}{\varphi^{2^n a + b} - (-1)^b \varphi^{-2^n a - b}} \right ) \\ &=\prod_{n=1}^{\infty} \left ( 1 + \frac{\varphi^{-2^n a - b} \left ( \varphi^b - (-1)^b \varphi^{-b} \right )}{1 - (-1)^b \varphi^{-2^{n+1}a - 2b}} \right ) \\ &=\prod_{n=1}^{\infty} \left ( 1 + \varphi^{-2^n a} \right ) \frac{1 - (-1)^b \varphi^{-2^n a - 2b}}{1 - (-1)^b \varphi^{-2^{n+1}a - 2b}} \\ &= \frac{1 - (-1)^b \varphi^{-2a-2b}}{1 - \varphi^{-2a}} \end{align*}$

by cancellations and the fact that $\displaystyle \prod_{n=1}^{\infty} \left ( 1 + x^{2^n} \right ) = \frac{1}{1-x^2} \;, \; \left | x \right | < 1$ . This completes the proof of the lemma.

Setting $a=b=2$ we get the result.

Telescopic Fibonacci product

January 6, 2023 Tolaso

Let $\mathcal{F}_n$ denote the Fibonacci sequence. Prove that

$\prod_{n=1}^{\infty} \frac{\mathcal{F}_{2n} \mathcal{F}_{2n+2} + \mathcal{F}_{2n-1} \mathcal{F}_{2n+2}}{\mathcal{F}_{2n} \mathcal{F}_{2n+2} + \mathcal{F}_{2n} \mathcal{F}_{2n+1}} = \sqrt{5}-1$

Solution

Let $P_N$ denote the partial product. Thus,

$\begin{align*} P_N &= \prod_{n=1}^{N} \frac{\mathcal{F}_{2n} \mathcal{F}_{2n+2} + \mathcal{F}_{2n-1} \mathcal{F}_{2n+2}}{\mathcal{F}_{2n} \mathcal{F}_{2n+2} + \mathcal{F}_{2n} \mathcal{F}_{2n+1}} \\ &=\prod_{n=1}^{N} \frac{\mathcal{F}_{2n+1} \mathcal{F}_{2n+2}}{\mathcal{F}_{2n} \mathcal{F}_{2n+3}} \\ &= \frac{\mathcal{F}_3 \mathcal{F}_4}{\mathcal{F}_2 \mathcal{F}_5} \cdot \frac{\mathcal{F}_5 \mathcal{F}_6}{\mathcal{F}_6 \mathcal{F}_7} \cdot \frac{\mathcal{F}_7 \mathcal{F}_8}{\mathcal{F}_6 \mathcal{F}_9} \cdots \frac{\mathcal{F}_{2N-1} \mathcal{F}_{2N}}{\mathcal{F}_{2N-2} \mathcal{F}_{2N+1}} \cdot \frac{\mathcal{F}_{2N+1} \mathcal{F}_{2N+2}}{\mathcal{F}_{2N} \mathcal{F}_{2N+3}} \\ &= \frac{\mathcal{F}_3}{\mathcal{F}_2} \cdot \frac{\mathcal{F}_{2N+2}}{\mathcal{F}_{2N+3}} \\ &= 2 \cdot \frac{\sqrt{5}-1}{2} \\ &= \sqrt{5}-1 \end{align*}$