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Arithmotheoretic sum

Evaluate the limit:

    \[\ell= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m\]

Solution

    \begin{align*} \ell &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} n \pmod m \\ &= \lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{m=1}^{n} \left ( n - m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=\lim_{n \rightarrow +\infty} \frac{1}{n^2} \left ( n^2 - \sum_{m=1}^{n} m \left \lfloor \frac{n}{m} \right \rfloor \right ) \\ &=1 - \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{m=1}^{n} \frac{m}{n} \left \lfloor \frac{n}{m} \right \rfloor \\ &= 1 - \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x \end{align*}

However,

    \begin{align*} \int_{0}^{1} x \left \lfloor \frac{1}{x} \right \rfloor \, \mathrm{d}x &\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{1}^{\infty} \frac{\left \lfloor u \right \rfloor}{u^3} \, \mathrm{d}x \\ &=\sum_{n=1}^{\infty} \int_{n}^{n+1} \frac{n}{x^3} \, \mathrm{d}x \\ &= \frac{1}{2}\sum_{n=1}^{\infty} n \left ( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{n}{(n+1)^2} \right )\\ &= \frac{1}{2} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{(n+1)^2} \right )\\ &=\frac{1}{2}\cancelto{1}{\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right )}+ \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(n+1)^2} \\ &=\frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \frac{1}{n^2} \\ &=\frac{1}{2} + \frac{1}{2} \left ( \frac{\pi^2}{6} - 1 \right ) \\ &=\frac{\pi^2}{12} \end{align*}

Hence

    \[\ell = 1 - \frac{\pi^2}{12}\]

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Series of Bessel function

Let J_n denote the Bessel function of the first kind. Prove that

    \[\sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 =1\]

Solution

The Jacobi – Anger expansion tells us that

(1)   \begin{equation*} e^{iz\sin\theta} = \sum_{n=-\infty}^{\infty}J_n(z) e^{in\theta}  \end{equation*}

Hence by Parseval’s Theorem it follows that

    \begin{align*} \sum_{n=-\infty}^{\infty}\left|J_n(z)\right|^2 &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\sin \theta} e^{iz\sin(-\theta)}\,\mathrm{d} \theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d} \theta \\ &=1 \end{align*}

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A factorial limit

Let \cdot! denote the factorial of a real number; that is x!=\Gamma(x+1). Evaluate the limit:

    \[\ell = \lim_{x \rightarrow n} \frac{x!-n!}{x-n}\]

Solution

It holds that

    \begin{align*} \lim_{x\rightarrow n} \frac{x!-n!}{x-n} &= \lim_{x\rightarrow n} \frac{\Gamma(x+1)- \Gamma(n+1)}{x-n} \\ &=\Gamma'(n+1) \\ &=\Gamma(n+1) \psi^{(0)}(n+1) \\ &=n! \left ( \mathcal{H}_n - \gamma \right ) \end{align*}

where \mathcal{H}_n denotes the n-th harmonic number and \gamma the Euler – Mascheroni constant.

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Linear isometry

Let f:\mathbb{R}^2 \rightarrow \mathbb{R}^2. If:

  • f(\mathbf{0})=\mathbf{0}
  • \left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right| for all {{\bf{u}},{\bf{v}}}

then prove that f is linear.

Solution

For convenience, identify \mathbb{R}^2 with \mathbb{C} here. Then note that for any such function f:\mathbb{C} \to \mathbb{C}, also z_1 \cdot f(z) a solution for any point z_1 on the unit circle. Also \overline{f(z)} is a solution. Note that \vert f(1)\vert=1 and hence we can wlog assume that f(1)=1. So f(i) is a point on the unit circle with distance \sqrt{2} to 1. Hence f(i) =\pm i, so w.l.o.g. assume that f(i)=i. But then for any z \in \mathbb{C}, both z and f(z) have the same distance to 0,1 and i. So supposing z \ne f(z), all 0,1,i lie on the perpendicular bisector between these points and in particular 0,1 and i are collinear which clearly is absurd. Hence f(z)=z for all z which proves the claim.

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Zeta logarithmic series

Let \zeta denote the zeta function. Prove that

    \[\sum_{k=1}^{\infty} \left( \log\left(1+\frac{1}{k}\right)-\frac{\zeta(k+1)}{k+1} \right)=0\]

Solution

    \begin{align*} \sum_{k=1}^{\infty} \left ( \frac{\zeta(k+1)}{k+1} - \frac{1}{k+1} \right ) &=\sum_{k= 2}^{\infty} \frac{\zeta(k)-1}{k}\\ &=\sum_{k =2}^{\infty} \sum_{n =2}^{\infty} \frac{1}{kn^k}\\ &=\sum_{n=2}^{\infty} \sum_{k= 2}^{\infty} \frac{1}{kn^k}\\ &=-\sum_{n=2}^{\infty} \left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)\\ &=-\sum_{n= 1}^{\infty} \left(\ln\left(1-\frac{1}{n+1}\right)+\frac{1}{n+1}\right) \\ &-\sum_{n =1}^{\infty} \left(\ln\left(\frac{n}{n+1}\right)+\frac{1}{n+1}\right)\\ &=\sum_{n= 1}^{\infty} \left(\ln\left(\frac{n+1}{n}\right)-\frac{1}{n+1}\right) \\ &=\sum_{n= 1}^{\infty} \left(\ln\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\right) \end{align*}

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