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Double summation

Evaluate the sum

    \[\mathcal{S} = \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-j^2}{\left ( n^2+j^2 \right )^2}\]

Solution

Lemma: It holds that

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

Proof: Consider the function \displaystyle  f(z)=\frac{\cot \pi z}{\sinh^2 \pi z} and let us it integrate over the following contour \gamma

By the residue theorem it follows that

    \[\oint \limits_{\gamma} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res}_{z_k} f(z)\]

For the residues we have

    \[\begin{matrix} \displaystyle \mathfrak{Res}_{z=n}{\frac{\cot\pi z}{\sinh^2\pi z}} &= & \displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=ni}{\frac{\cot\pi z}{\sinh^2\pi z}} &= &\displaystyle \frac{1}{\pi\sinh^2\pi n} \\\\ \displaystyle \mathfrak{Res}_{z=0}{\frac{\cot\pi z}{\sinh^2\pi z}} & = &\displaystyle -\frac{2}{3\pi} \end{matrix}\]

The integrals along the sides vanish; hence:

    \[-2i\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=2\pi i\left(-\frac{2}{3\pi}+\frac{4}{\pi}\sum_{n=1}^\infty\frac{1}{\sinh^2\pi n}\right)\]

and since

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2\pi x}=\left[ \frac{1}{\pi}\tanh\pi x \right]_{-\infty}^\infty=\frac{2}{\pi}\]

the result follows. \blacksquare

Back to the problem. We have successively:

    \begin{align*} \mathcal{S} &= \frac{1}{2} \sum_{j=1}^{\infty} \sum_{n=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2}\sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \prod_{n=1}^{\infty} \left ( 1 + \frac{j^2}{n^2} \right ) \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \frac{\mathrm{d}^2 }{\mathrm{d}^2 j} \ln \frac{\sinh \pi j }{\pi j} \\ &=\frac{1}{2} \sum_{j=1}^{\infty} \left ( \frac{1}{j^2} - \frac{\pi^2}{\sinh^2 \pi j} \right ) \\ &= \frac{\pi}{4} \end{align*}

 

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Arithmotheoretic sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^\infty\frac{1}{n}\sum_{d\mid n} \frac{d}{n+d^2}\]

Solution

The sum converges absolutely , so we can switch the order of summation; hence:

    \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \sum_{d\mid n} \frac{d}{n+d^2} &= \sum_{d=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{dk} \frac{d}{dk+d^2} \\ &= \sum_{d=1}^{\infty} \frac{1}{d} \sum_{k=1}^{\infty} \frac{1}{k(d+k)} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^2} \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{d+k} \right) \\ &= \sum_{d=1}^{\infty} \frac{\mathcal{H}_d}{d^2} \end{align*}

The last sum equals 2\zeta(3) and hence

    \[\mathcal{S} = 2\zeta(3)\]

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Gamma inequality

Let a ,b, c be three positive real numbers such that abc=1. Prove that:

    \[\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{\Gamma(b)}{1+b+bc} + \frac{\Gamma(c)}{1+c+ca} \geq 1\]

where \Gamma is Euler’s Gamma function.

Solution

We can rewrite the inequality as

    \begin{align*} \sum \frac{\Gamma\left ( a \right )}{1+a+ab} &= \frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a}{a} \frac{\Gamma(b)}{1+b+bc} + \frac{ab}{ab} \frac{\Gamma(c)}{1+c+ca} \\ &=\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a\Gamma(b)}{a+ab+abc} + \frac{ab\Gamma(c)}{ab+abc+abca} \\ &=\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a \Gamma(b)}{1+a+ab} + \frac{ab \Gamma(c)}{1+a+ab} \\ &=\frac{\Gamma(a) + a \Gamma(b) + ab \Gamma(c)}{1+a+ab} \\ &\geq \Gamma \left ( \frac{1+a + ab}{1+a+ab} \right ) \\ &= 1 \end{align*}

since \Gamma is convex.

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Asymptotic expansion

Prove that

    \[\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{i^2+j^2} = \frac{\pi \log n}{2} + \frac{\pi+4}{4n} + c + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

for some constant c.

A product

Let \phi denote the golden ratio , \mu the Möbius function and \varphi  Euler’s totient function. Prove that:

    \[\prod_{n=1}^{\infty} \left ( 1 - \frac{1}{\phi^n} \right )^{\frac{\mu(n) - \varphi(n)}{n}} = e\]

Solution

We are using the following facts:

(1)   \begin{equation*} \sum_{d \mid m} \varphi(d) = m \end{equation*}

(2)   \begin{equation*} \sum_{d \mid m} \mu(d) = \left\{\begin{matrix} 1 & , & m=1 \\ 0 & , & m>1 \end{matrix}\right. \end{equation*}

Hence,

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mu(n) - \varphi(n)}{n}\log\left(1 - \frac{1}{\varphi^n} \right) &=-\sum_{n=1}^{\infty}\frac{\mu(n) - \varphi(n)}{n} \sum_{k=1}^{\infty} \frac{1}{k\varphi^{kn}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m \varphi^m} \sum_{d|m} \left(\varphi(d) - \mu(d) \right) \\ &= \sum_{m=2}^{\infty} \frac{1}{\varphi^m}\\ &= \frac{1}{\varphi^2} \frac{1}{1- \frac{1}{\varphi}}\\ &= \frac{1}{\varphi^2-\varphi} \\ &= 1 \end{align*}

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