An interesting infinite product!

\[\displaystyle\prod_{n=1}^\infty\coth^{(-1)^n}\left(\dfrac{\pi n}{2}\right)=\dfrac{\sqrt\pi}{\sqrt[4]2\sqrt{\varpi}}\]

where \(\varpi=2.62205755\ldots\) is the lemniscate constant (the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of \(\pi\) for the circle). Both \(\varpi\) and \(\pi\) are proven to be transcendental.

#lemniscate #lemniscateconstant #product #infiniteproduct #HyperbolicFunction #HyperbolicCotangent #Function #pi #maths #math

For any integer \(n\geq3\),

\[\left\lfloor\dfrac{1}{1-\displaystyle\prod_{k=n}^\infty\left(1-\dfrac{1}{F_k}\right)}\right\rfloor=F_{n-2}\]
where \(F_n\) is the \(n\)-th Fibonacci number.
#FibonacciNumber #Fibonacci #FibonacciSequence #FibonacciSeries #Product #FloorFunction #Floor #InfiniteProduct

Interestingly, \(\pi^\text{th}\) root of \(4\) can be expressed as the following limit.
\[\boxed{\displaystyle\sqrt[\pi]4=\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}}\]

Proof:
Let \(\Phi_n=\displaystyle\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}\). And taking natural logarithms on both sides, we get
\[\begin{align*}\ln\Phi_n&=\displaystyle\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\tan^{-1}(n+k)\right)\\&=\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\left(\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\right)\\&=-\sum_{k=0}^n\ln\left(1-\dfrac{2}{\pi}\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\\&=-\sum_{k=0}^n\left(-\dfrac{2}{\pi(n+k)}+O\left(\dfrac{1}{n^2}\right)\right)\\&=\dfrac{2}{\pi}\sum_{k=0}^n\dfrac{1}{n+k}+O\left(\dfrac{1}{n}\right)\end{align*}\]
Taking limit \(n\to\infty\), we get
\[\begin{align*}\displaystyle\lim_{n\to\infty}\ln\Phi_n&=\dfrac{2}{\pi}\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=0}^n\dfrac{1}{1+\frac{k}{n}}+\lim_{n\to\infty}O\left(\dfrac{1}{n}\right)\\&=\dfrac{2}{\pi}\displaystyle\int_0^1\dfrac{\mathrm{d}x}{1+x}\\&=\dfrac{2}{\pi}\ln2=\ln\sqrt[\pi]4\end{align*}\]

Hence, the required limit:
\[\boxed{\displaystyle\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}=\displaystyle\lim_{n\to\infty}\Phi_n=\sqrt[\pi]4}\ \textbf{Q.E.D.}\ \blacksquare\]

*Note:
\(\tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+O(x^6)\)
\(\ln(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dfrac{x^5}{5}+O(x^6)\)
\(\ln\left(1-\dfrac{2}{\pi}\tan^{-1}(x)\right)=-\dfrac{2x}{\pi}+O(x^2)\)

#pi #limit #product #proof #pithrootof4 #interestinglimit #infiniteproduct #infinitesum #logarithm

PRODUCTS OVER PRIME NUMBERS [2/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p+1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-3}}{1-p^{-2}}=\dfrac{\zeta(2)}{\zeta(3)}=\dfrac{\pi^2}{6\zeta(3)}\]
\[\displaystyle\prod_{p\in\mathbb{P}}\left(1+\dfrac{1}{p(p-1)}\right)=\prod_{p\in\mathbb{P}}\dfrac{1-p^{-6}}{(1-p^{-2})(1-p^{-3})}=\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}=\dfrac{315}{2\pi^4}\zeta(3)\]
#PrimeProducts #RiemannZetaFunction #EulerProduct #ZetaFunction #InfiniteProduct #NumberTheory

SOME PRODUCTS OVER PRIME NUMBERS [1/2]:
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}=\prod_{p\in\mathbb{P}}\left(\sum_{k=0}^\infty\dfrac{1}{p^{ks}}\right)=\sum_{n=1}^\infty\dfrac{1}{n^s}=\zeta(s)\]
\[\displaystyle\prod_{p\in\mathbb{P}}\dfrac{p^s+1}{p^s-1}=\prod_{p\in\mathbb{P}}\left(1+\frac{2}{p^s}+\frac{2}{p^{2s}}+\cdots\right)=\sum_{n=1}^\infty\dfrac{2^{\omega(n)}}{n^s}=\dfrac{\zeta(s)^2}{\zeta(2s)}\]
#PrimeProducts #EulerProduct #RiemannZetaFunction #ZetaFunction #InfiniteProduct