Joy of Mathematics

A Fibonacci series

July 17, 2022July 24, 2022 Tolaso

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

$\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n-1}}{\mathcal{F}_n \mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+2}}{\mathcal{F}_n \mathcal{F}_{n+1}-1}\right)=\frac{\pi^2}{16}$

Solution

Let $S$ denote the given sum. Then,

$\begin{align*} \mathcal{S} & =\sum_{n=2}^{\infty}\arctan \left(\frac{\mathcal{F}_{n+1}-\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}+1}\right)\arctan \left(\frac{\mathcal{F}_{n+1}+\mathcal{F}_n}{\mathcal{F}_n\mathcal{F}_{n+1}-1}\right) \\ &=\sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)(\pi-\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n) \\ &= \pi \sum_{n=2}^{\infty}(\arctan \mathcal{F}_{n+1}-\arctan \mathcal{F}_n)- \sum_{n=2}^{\infty}\left(\arctan^2 \mathcal{F}_{n+1} -\arctan^2 \mathcal{F}_n\right) \\ &= \pi \lim_{n \rightarrow +\infty}(\arctan \mathcal{F}_n-\arctan \mathcal{F}_2)- \lim_{n \rightarrow + \infty} \left( \arctan^2 \mathcal{F}_n -\arctan^2 \mathcal{F}_2 \right) \\ &= \pi \left ( \frac{\pi}{2} - \frac{\pi}{4} \right ) - \left ( \frac{\pi^2}{4} - \frac{\pi^2}{16} \right ) \\ &= \frac{\pi^2}{16} \end{align*}$

A logarithmic integral

May 14, 2022 Tolaso

Let $\alpha \in \mathbb{R}$ . Prove that:

$\int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2}\, \mathrm{d}x = \frac{\pi}{\sqrt{2}} \cdot \frac{\alpha}{\sqrt{1+ \sqrt{1+\alpha^2}}}$

Solution

Using the known Fourier series $\displaystyle \sum_{n=-\infty}^{\infty} \frac{1}{\left( x+2\pi n \right)^2}=\frac{1}{4} \csc^2 \frac{x}{2}$ as well as the known generating function of the Catalan numbers

$\sum_{n=0}^{\infty} \mathcal{C}_n x^n = \frac{1- \sqrt{1-4x}}{2x} \Rightarrow \sum_{n=0}^{\infty} \mathcal{C}_{2n} x^{2n} = \frac{\sqrt{1+4x} - \sqrt{1-4x}}{4x}$

Then, we have successively:

$\begin{align*} \int_{0}^{\infty} \frac{\arctan \alpha \sin^2 x}{x^2} \, \mathrm{d}x &=\frac{1}{2} \int_{-\infty}^{\infty}\frac{\arctan \left(\alpha \sin^{2} x\right)}{x^{2}}\, \mathrm{d}x\\ &=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan \left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}} \, \mathrm{d}x\\ &=\frac{1}{8}\int_{0}^{2\pi} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}\frac{x}{2} \, \mathrm{d}x\\ &=\frac{1}{4}\int_{0}^{\pi/2}\arctan \left(\alpha \sin^{2} x \right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right) \, \mathrm{d} x\\ &=\int_{0}^{\pi/2} \arctan\left(\alpha \sin^{2}x\right)\csc^{2}x\, \mathrm{d} x\\ &=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n}}{2n+1} \alpha^{2n+1} \int_{0}^{\pi/2}\sin^{4n}x \, \mathrm{d}x\\ &=\pi \alpha \sum_{n=0}^{\infty} \mathcal{C}_{2n}\left(-\frac{\alpha^{2}}{16}\right)^{n} \\ &=\frac{\pi}{i}\left(\sqrt{1+i \alpha}-\sqrt{1-i\alpha}\right)\\ &=\frac{\pi}{\sqrt{2}}\frac{\alpha}{\sqrt{1+\sqrt{1+\alpha^{2}}}} \end{align*}$

A Gamma summation

May 11, 2022 Tolaso

Let $a \notin \mathbb{Z}$ and $a > \frac{1}{2}$ . Prove that

$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$

Solution

Consider the function

$f(x) = \left\{\begin{matrix} \cos^{2a-2} \pi x & , & \left | x \right |< \frac{1}{2} \\ 0 & , & \text{elsewhere} \end{matrix}\right.$

We easily evaluate that

$\hat{f}(y)= \int_{-\infty}^{\infty} f(x) e^{-2i \pi x y} \, \mathrm{d}x = \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+y) \Gamma(a-y)}$

Hence it follows from Poisson summation formula that

$\begin{align*} 1 &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) \\ &= \sum_{n=-\infty}^{\infty} \frac{2^{2-2a} \Gamma(2a-1)}{\Gamma(a+n) \Gamma(a-n)} \end{align*}$

Thus,

$\begin{align*} \frac{1}{2^{2-2a} \Gamma(2a-1)} &= \sum_{n=-\infty}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &=\sum_{n=-\infty}^{-1} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(\alpha)} + \\ &\quad \quad \quad + \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} \\ &= 2 \sum_{n=1}^{\infty} \frac{1}{\Gamma(a+n) \Gamma(a-n)} + \frac{1}{\Gamma^2(a)} \\ &= 2 \sum_{n=0}^{\infty}\frac{1}{\Gamma(a+n) \Gamma(a-n)} - \frac{1}{\Gamma^2(a)} \end{align*}$

The result follows.

A rational harmonic series

April 26, 2022 Tolaso

Can the rational numbers in the interval $[0,1]$ be enumerated as a sequence $\{q_n\}_{n \geq 1}$ in such a way that the series $\displaystyle \sum_{n=1}^{\infty} \frac{q_n}{n}$ is convergent?

Infinite product

April 20, 2021 Tolaso

Prove that

$\prod_{n=0}^{\infty} \frac{\cosh \left ( n^2 + n + \frac{1}{2} \right ) + i \sinh \left ( n + \frac{1}{2} \right )}{\cosh \left ( n^2 + n + \frac{1}{2} \right ) - i \sinh \left ( n + \frac{1}{2} \right )} = i$