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An series related to the eta Dedekind function

Prove that

    \[\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi} = \frac{1}{6} - \frac{1}{2\pi }\]

Solution

Let us consider the function

    \[f(z) = \frac{\cot \pi  z}{\sinh^2 \pi z}\]

and integrate it along a quadratic counterclockwise contour \Gamma_N with vertices \displaystyle \frac{N}{2} \left ( \pm 1 \pm i \right ) where N is a big odd natural number. Hence,

    \begin{align*} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \int_{N/2}^{-N/2} f \left ( x + \frac{i N}{2} \right )\, \mathrm{d}x + i \int_{N/2}^{-N/2} f \left ( i y + \frac{N}{2} \right ) \, \mathrm{d}y \\ &\quad \quad \quad + \int_{-N/2}^{N/2} f \left ( x - \frac{i N}{2} \right ) \, \mathrm{d}x + i \int_{-N/2}^{N/2} f \left ( iy + \frac{N}{2} \right ) \, \mathrm{d}y \end{align*}

We note that \displaystyle \lim_{N \rightarrow +\infty} f \left (x \pm \frac{i N}{2} \right ) = \frac{\mp i}{\cosh^2 \pi x} and that \displaystyle \lim_{N \rightarrow +\infty} f \left (i y \pm \frac{N}{2} \right ) = 0.

It’s also easy to see that

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2 \pi x } = \left [ \frac{\tanh \pi x}{\pi} \right ]_{-\infty}^{\infty} = \frac{2}{\pi}\]

Hence, as N \rightarrow +\infty we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = - \frac{4 i}{\pi}\]

By Residue theorem we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res} \left ( f; z = k \right ) + \mathfrak{Res} \left ( f; z = ik \right )\]

It is straightforward to show that

    \[\begin{matrix} \mathfrak{Res} \left ( f;z=k \right ) = &\mathfrak{Res} \left ( f;z=ik \right ) = & \dfrac{1}{\pi \sinh^2 \pi k} \\\\ \mathfrak{Res} \left ( f;z=0 \right ) & = & -\dfrac{2}{3\pi} \end{matrix}\]

Hence,

    \[\oint \limits_{\Gamma_N} f(z)\, \mathrm{d}z = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2 n \pi}-\frac{4 i}{3}\]

in the limit N \rightarrow +\infty. The result follows.

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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.

Solution

Let p_k be the k-th prime number.

    \[\sum_{n\ge1}\frac{1}{n\operatorname{gpf}(n)}=\sum_{k \geq 1}\frac{1}{p_k}\sum_{\operatorname{gpf}(n)=p_k} \frac{1}{n}\]

If \operatorname{gpf}(n)=p_k , then n=p_1^{i_1}\dots p_{k-1}^{i_{k-1}}\,p_k^{i_k} with i_j \geq 0, 1 \leq j<k and i_k \geq 1. It folows that

    \begin{align*} \sum_{\operatorname{gpf}(n)=p_k}\frac{1}{n}&=\left(\sum_{i_1 \geq 0}p_1^{-i_1} \right)\cdots \left(\sum_{i_{k-1}\geq 0} p_{k-1}^{-i_{k-1}} \right) \left(\sum_{i_k\ge1}p_k^{-i_k}\right)\\ &=\frac{1}{p_k}\,\prod_{i=1}^k \left(1-\frac{1}{p_i} \right)^{-1} \end{align*}

From Merten’s theorem

    \[\prod_{i=1}^k \left(1-\frac{1}{p_i} \right)^{-1}\sim e^\gamma\,\log p_k\]

and the original series has the same character as

    \[\sum_{k=1}^{\infty} \frac{\log p_k}{p_k^2}\]

which is convergent.

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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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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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