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


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

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It follows from Taylor's theorem that $f(z)=\sum \limits_{n=0}^{\infty} c_n z^n$ and that the convergence is uniform. Thus, \begin{align*} \frac{1}{2\pi i }\oint \limits_{|z|=1} \frac{\overline{f(z)}}{z-\alpha} \,\mathrm{d}z &=\frac{1}{2\pi i }\oint \limits_{|z|=1} \sum_{n=0}^{\infty} \frac{\ove...

It holds that $${\rm nul} (T_1 T_2) \leq {\rm nul} (T_1) + {\rm nul} (T_2)$$ where $T_1, \; T_2$ are the corresponding linear transformations. Proof: The proof of the lemma is based on the rank - nullity theorem. Based upon the above lemma we have that \begin{align*} {\rm rank} \left ( T_1 T_2 \rig...

The sum of $D(\sigma)$ over the even permutations minus the one over the odd permutations is the determinant of the matrix $A$ with entries $a_{i,j}=\vert i-j\vert$ and this determinant is known to be

$$\det A = (-1)^{n-1} (n-1) 2^{n-2}$$

Let $A, B$ be elements of an arbitrary associative algebra with unit. Then: \begin{align*} \left ( A^{-1} +\left ( B^{-1} - A \right )^{-1} \right )^{-1} &= \left ( A^{-1} \left ( B^{-1} - A \right )\left ( B^{-1} - A \right )^{-1} + A^{-1} A \left ( B^{-1} - A \right )^{-1} \right )^{-1} \\ &am...

Let us suppose that $|\mathcal{G}| = \kappa$ and $x= \frac{1}{\kappa} \sum \limits_{g \in \mathcal{G}} g $. We note that for every $h \in \mathcal{G}$ the depiction $\varphi: \mathcal{G} \rightarrow \mathcal{G}$ such that $\varphi(g)=h g $ is $1-1$ and onto. Thus: \begin{align*} x^2 &=\left ( \f...

Using  $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of  $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$. Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$. Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n...

Hint: Equality holds when vectors are parallel i.e, $u=kv$, $k \in \mathbb{R}^+$ because $u \cdot v= \|u \| \cdot \|v\| \cos \theta$ when $\cos \theta=1$, the equality of the Cauchy-Schwarz inequality holds.