Product of sines

$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$$

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Modification on a sequence from VJIMC

The first two limits of the following problem were proposed at VJIMC, 2005, Category I. The exact value of the last limit was proposed by a user at https://artofproblemsolving.com , where you can also see his solution along other lines.  Let $(x_n)_{n\ge 2}$ be a sequence of real numbers, such that $x_2>0$ and for every $n\ge 2$ holds  $$x_{n+1}=-1+\sqrt[n]{1+nx_n}.$$ Prove consecutively that  $$1)\ \lim_{n\to\infty}x_n=0,\  \ \ 2)\ \lim_{n\to\infty}nx_n=0,\  \ \ 3)\ \lim_{n\to\infty}n^2x_n=4.$$ $\textbf{Proof.}$  $\textbf{1)}$ Clearly all the elements of the sequence are positive. The inequality $-1+\sqrt[n]{1+nx_n}<x_n$ is equivalent to $(1+nx_n)<(1+x_n)^n$, which is seen to be true after expanding, since all the summands on the right are positive. This shows that the sequence is strictly decreasing. Hence $$0<x_{n+1}=-1+\sqrt[n]{1+nx_n}\le -1+\sqrt[n]{1+nx_2},$$ and since the right hand side clearly tends to $0$ we obtain $\displaystyle\lim_...
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OMOUS 2024

Let $A$ be a real matrix such that $A+A^2A^t+(A^t)^2=0$. Prove that $A=0$. $\textbf{Solution}.$ Multiply the given equation by $A^t$ from the right to get $AA^t+A^2(A^2)^t+(A^t)^3=0$ and hence $(A^t)^3=-AA^t-A^2(A^2)^t$ . Observe that RHS is negative semi-definite symmetric matrix. Thus $(A^t)^3$ and hence $A^3$ is symmetric negative semi-definite. Now multiply the given equation with $A$ from the left to get  $A^2+A^3A^t+A(A^t)^2=0$ and using $A^3=(A^t)^3$ we get $A^2+(A^t)^4+A(A^t)^2=0$. Transposing this equation we get $(A^t)^2+A^4+A^2 A^t=0$. Comparing with the original equation we get $A^4=A$. Thus $A$ is diagonalizable and $\lambda^4=\lambda$ for every characteristic root $\lambda$ of $A$. Thus  $\lambda=0$ or $\lambda^3=1$. But $A^3$ is negative semi-definite, so $\lambda^3\le 0$. Thus $\lambda=0$ and hence all roots of $A$ are $0$. Since $A$ is diagonalizable, $A=0$.
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Proof of Gelfand-Phillips Theorem

$\textbf{Theorem.}$ Let $X$ be a Banach space and $A\subseteq X$. Prove that $A$ is precompact (in the norm) if and only if for every $w^*$-convergent to $\textbf{0}$ sequence $\{x_n^*\}_{n\ge 1}\subseteq X^*$ it holds that $\{x_n^*\}_{n\ge 1}$ converges uniformly to $\textbf{0}$ on  $A.$   $\textbf{Proof.}$ If $A$ is precompact the result follows from Arzela-Ascoli theorem (combined with the fact that $w^*$-convergent sequences are norm bounded). Now we prove the reverse direction.  First we proof the following characterization of compactness in normed spaces: $A$ is compact if and only if $A$ is bounded and for any $\varepsilon>0$ there exists a finite-dimensional space $F$ such that $A\subseteq F+\varepsilon\mathbf{B}.$ $\textit{Proof}.$  If $A$ is precompact, then for any $\varepsilon>0$ there exists $\{a_i\}_{i=1}^n\subseteq A$ such that $A\subseteq\bigcup_{i=1}^n\mathbf{B}_{\varepsilon}(a_i)$. In particular $A$ is bounded and \[A\subseteq \text{span}(\{...
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