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Justify the fact that:

$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \rightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$

Evaluate in a closed form the series:

$$\mathcal{S} = \sum_{n=-\infty}^{\infty} \frac{16}{\left ( 4n^2+n+5 \right )^2}$$

First of all note that $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ hence $10!$ is not a perfect square, $10!$ has $(8+1)(4+1)(2+1)(1+1) = 270$ divisors. If $d \mid 10!$ , then there exists $p$ such that $dp =10!$ meaning that $p$ is also a divisor of $10!$. We also note that if one of $p, q$ is less tha...

Endow $(0, +\infty)$ with the following metric

$$d(x, y) = \left | \log \frac{x}{y} \right |$$

(i) Verify that $d$ is indeed a metric.

(ii) Is the sequence $a_n =\frac{1}{n}$ convergent under this metric? Give a brief explanation.

(iii) Examine if $(0, +\infty)$ is bounded under this metric.

Hi Ram_1729. Could you explain what $Q_8$ and $K_4$ stand for?

Thank you Ram_1729. Exactly! It was an exam's question!

Let $x_1, x_2, \dots, x_n$ be $n \geq 2$ positive numbers other than $1$ such that $x_1^2+x_2^2+\cdots +x_n^2=n^3$. Prove that:

$$\frac{\log_{x_1}^4 x_2}{x_1+x_2}+ \frac{\log_{x_2}^4 x_3}{x_2+x_3}+ \cdots + \frac{\log_{x_n}^4 x_1}{x_n+x_1} \geq \frac{1}{2}$$

The tetrahedron is a regular solid with $4$ vertices and $4$ triangular faces. The symmetry group is the alternating group $\mathcal{A}_4$. The symmetry group of a cube is isomorphic to $\mathcal{S}_4$ , the permutation group on 4 elements. If we number the vertices of the cube from $1$ to $4$ and ...

Let $A, B \in \mathcal{M}_{2 \times 2}$ be matrices with integer entries such that $AB = BA$ , $\det \left( A + B \right) =2$ and $\det \left( A^3 + B^3 \right) = 2^3$. Evaluate the determinant


$$\mathcal{D} = \det \left( A^2 + B^2 \right)$$

Let $\mathcal{H}_n$ denote the $n$ - th harmonic number. It holds that $$\sum\limits_{n=1}^{\infty}\mathcal{H}_{pn}x^n = -\frac{1}{p}\sum\limits_{k=0}^{p-1} \frac{\ln \varphi_k}{\varphi_k}$$ where $p \in \mathbb{N}$ and $\displaystyle \varphi_k = \varphi_k(x) = 1 - \sqrt[p]{x}\exp\left(\frac{-2\pi i...