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Double summation

Evaluate the sum

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


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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On functions

  1. Does there exist an entire f:\mathbb{C} \rightarrow \mathbb{C} such that f \left( \frac{1}{n} \right) =\frac{1}{n} for all n \in \mathbb{N}?
  2. Does there exist a holomorphic function f on the unit disk such that f \left( \frac{1}{n} \right) =\frac{1}{n!} for all n \in \mathbb{N}?


  1. Yes. It follows immediately from the identity theorem.
  2. Νο, there is no such function. If there was, its Taylor series centered at 0 would be of the form

        \[f(z) = a_kz^k+a_{k+1}z^{k+1}+\cdots\]

    for some k \in \mathbb{N} and a_k \neq 0. But then

        \[\lim_{n \rightarrow + \infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\left\lvert\frac{a_k}{n^k}\right\rvert}=1\]

    In particular

        \[\lim_{n \rightarrow + \infty}\frac{\left\lvert f\left(\frac1n\right)\right\rvert}{\frac1{n^k}} \neq 0\]


        \[\lim_{n \rightarrow +\infty}\frac{\frac1{n!}}{\frac1{n^k}}=0\]

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Explicit formula of f

In the following figure the function f is continuous and 1-1. For every point \mathrm{P} on the curve y=2x^2 the areas of \mathrm{A} and \mathrm{B} are equal.

Rendered by QuickLaTeX.com

You are asked to find an explicit formula for f.


We are expanding the above figure so that it looks like this;

Rendered by QuickLaTeX.com

The area \mathrm{A} is given by

    \[\mathrm{A} = \int_{0}^{\mathbf{b}}2x^2 \, \mathrm{d} x - \int_{0}^{\mathbf{b}} x^2 \, \mathrm{d}x = \int_{0}^{\mathbf{b}} x^2 \, \mathrm{d}x = \frac{\mathbf{b}^3}{3}}\]

On the other hand we have that

    \begin{align*} \mathbf{b} f(a) &= \int_{0}^{\mathbf{b}} x^2 \, \mathrm{d}x + \mathrm{A} + \mathrm{B} + \left ( a f(a) - \int_{0}^{\mathbf{a}} f(x) \, \mathrm{d}x \right ) \\ &\!\!\!\!\!\!\!\!\!\!\!\!\overset{\mathrm{A} = \mathrm{B} \; , \; f(a)=2b^2}{=\! =\! =\! =\! =\! =\!=\! =\!=\! =\!} \int_{0}^{\mathbf{a}} f(x) \, \mathrm{d}x = 2\mathbf{a} \mathbf{b}^2 - \mathbf{b^3} \end{align*}

Differentiating with b as a variable and a(b) dependent to b we get that \mathbf{b} = \frac{3a}{4}. Hence,

    \[f(x) = \frac{32x^2}{9}\]

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Limit of a Riemann sum

Let f:[0,1] \rightarrow \mathbb{R} be \mathcal{C}^1. Prove that

    \[\lim_{n \rightarrow +\infty} n \left ( \frac{1}{n} \sum_{k=1}^{n} f \left ( \frac{k}{n} \right ) - \int_{0}^{1} f(x) \, \mathrm{d}x \right ) = \frac{f(1)-f(0)}{2}\]

Arithmotheoretic sum

Evaluate the sum

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


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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