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.

Double inequality

In a triangle ABC prove that

    \[\frac{4}{9} \sum \sin B \sin C \leq \prod \cos \frac{B-C}{2} \leq \frac{2}{3} \sum \cos A\]

Trigonometric series

Let a \in \mathbb{R}. Prove that

    \[\sum_{n=1}^\infty 2^{2n}\sin^4 \frac a{2^n}=a^2-\sin^2a\]

Solution

First of all we note that

    \begin{align*} 2^{2n}\sin^4\frac a{2^n}&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\sin^2\frac a{2^n}\\ &=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\left(1-\cos^2\frac a{2^n}\right)\\ &=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n}\cdot\sin^2\frac a{2^n}\cos^2\frac a{2^n}\\ &=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}} \end{align*}

Hence,

    \begin{align*} \sum_{n=1}^{m}2^{2n}\sin^4 \frac{a}{2^n} &= \sum_{n=1}^{m} \left( 2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}} \right) \\ &= 2^{2m}\sin^2\frac{a}{2^m}-\sin^2a \end{align*}

Letting m \rightarrow +\infty we get the requested value.

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