Triangle relation

Prove that in any triangle \mathrm{AB} \Gamma it holds that

    \[\left ( \frac{\beta}{\gamma} + \frac{\gamma}{\beta} \right ) \cos \mathrm{A} + \left ( \frac{\gamma}{\alpha} + \frac{\alpha}{\gamma} \right ) \cos \mathrm{B} + \left ( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right ) \cos \Gamma = 3\]

Convergence of series

Let \mathrm{gpf} denote the greatest prime factor of n. For example \mathrm{gpf}(17) = 17 , \mathrm{gpf} (18) =3. Define \mathrm{gpf}(1)=1. Examine if the sum

    \[\mathcal{S} = \sum_{n=1}^{\infty} \frac{1}{n \; \mathrm{gpf}(n)}\]

converges.

On a prime summation

Let p_n denote the n – th prime. Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^{\infty} \frac{ \log p_n}{p_n^2 -1}\]

Square summable

Let \{a_n\}_{n \in \mathbb{N}} be a real sequence such that

If \{b_n\}_{n \in \mathbb{N}} is a real sequence that is square summable; i.e \sum \limits_{n=1}^{\infty} b_n^2 < +\infty the sequence \sum \limits_{n=1}^{\infty} a_n b_n converges.

Prove that \{a_n\}_{n \in \mathbb{N}} is also square summable.

Solution

Let f_N:\ell_2 \rightarrow \mathbb{R} be defined as

    \[f_N(b)=\sum_{n=1}^{N} a_nb_n\]

where b = (b_n) \in \ell_2. We note that

    \begin{align*} |f_N(b)|^2 &\leq \sum_{n=1}^N|a_n|^2 \sum_{n=1}^N|b_n|^2 \\ & \leq ||b||^2\sum_{n=1}^N|a_n|^2 \end{align*}

Equality holds when b=(a_1, \dots, a_N, 0, 0, \dots ). Hence, \displaystyle ||f_N||= \left( \sum_{n=1}^N|a_n|^2\right) ^{1/2}. From the hypothesis, it follows that \{f_n\}_{n \in \mathbb{N}} is pointwise bounded. It follows from the Uniform boundedness principle ( Banach – Steinhaus ) that \displaystyle ||f_N||= \left( \sum_{n=1}^N|a_n|^2\right) ^{1/2} are bounded. Hence, \{a_n\}_{n \in \mathbb{N}} is square summable.

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The integral domain is a field

Prove that an integral domain with the property that every strictly decreasing chain of ideals must be finite in length is a field.

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