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A root limit

Let a_i \; , \; i =1, 2, \dots, k be positive real numbers such that a_1\geq a_2\geq \cdots \geq a_k. Prove that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} = \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

Solution

Without loss of generation , let a_1 =\max\{ a_1, a_2 , \dots , a_k \}. Then,

    \begin{align*} a_1 &=(a_1^n)^{1/n}\\ &\leq (a_1^n+\cdots+a_k^n)^{1/n}\\ &=\sqrt[n]{a_1^n \left (1+\left (\frac{a_2}{a_1} \right )^n + ... + \left (\frac{a_k}{a_1} \right )^n \right )} \\ &\leq \sqrt[n]{{a_1}^n \cdot k} \\ &= a_1\sqrt[n]{k} \\ &=a_1 k^{1/n} \end{align*}

since a_i \leq  a_1 forall i=1,2, \dots,k. Thus, by the squeeze theorem it follows that

    \[\lim_{n \rightarrow +\infty} \sqrt[n]{a_1^n + a_2^n + \cdots + a_k^n} =a_1 =  \max\left \{ a_1, a_2 , \dots, a_k \right \}\]

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A limit!

Evaluate the limit:

    \[\ell = \lim_{m\rightarrow +\infty}\sqrt[m]{m+1}\cdot\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}\]

Solution

Let a_m=\sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}}. Then,

    \[\log a_m = \log \sqrt[m^2+m]{\binom{m}{1}\binom{m}{2}\cdots\binom{m}{m}} = \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k}\]

It follows by Stolz–Cesàro that

    \begin{align*} \lim_{m \rightarrow +\infty} \log a_m &= \lim_{m \rightarrow +\infty} \frac{1}{m^2+m} \sum_{k=1}^{m} \log \binom{m}{k} \\ &=\lim_{m \rightarrow +\infty} \frac{\sum \limits_{k=1}^{m+1} \log \binom{m+1}{k} - \sum \limits_{k=1}^{m} \log \binom{m}{k}}{(m+1)^2 + (m+1) -(m^2+m)} \\ &=\lim_{m \rightarrow +\infty} \frac{1}{2m+2} \sum_{k=1}^{m} \log \frac{m+1}{m+1-k} \\ &= \lim_{m \rightarrow +\infty} \frac{\log \frac{(m+1)^m}{m!}}{2m+2}\\ &=\lim_{m \rightarrow +\infty} \frac{1}{2} \cdot \frac{m}{m+1} \cdot \log \frac{m+1}{\sqrt[m]{m!}} \\ &= \frac{1}{2} \cdot 1 \cdot \log e \\ &= \frac{1}{2} \end{align*}

In addition,

    \begin{align*} \lim_{m \rightarrow +\infty} \sqrt[m]{m+1} &= \lim_{m \rightarrow +\infty} \left ( m+1 \right )^{1/m} \\ &=\lim_{m \rightarrow +\infty} e^{\frac{\log m}{m+1}} \\ &=1 \end{align*}

Hence \ell = \sqrt{e}.

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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]

Solution

We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}

Hence,

    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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Logarithmic Gaussian integral

Let \gamma denote the Euler’s constant. Prove that:

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = -\frac{\sqrt{\pi}}{4} \left ( 2 \log 2 + \gamma \right )\]

Solution

First of all by making the substitution x \mapsto \sqrt{x} we get that

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = \frac{1}{4} \int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x\]

For t>-1 let us consider the function

    \[F(t) = \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}x = \Gamma\left ( t+1 \right )\]

where \Gamma is the Euler’s Gamma function. Differentiating once we get:

    \begin{align*} \Gamma'(t+1) &=\frac{\mathrm{d} }{\mathrm{d} t} \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}t \\ &=\int_{0}^{\infty} \frac{\partial }{\partial t} x^t e^{-x}\; \mathrm{d}t \\ &=\int_{0}^{\infty} x^t \log x e^{-x} \; \mathrm{d}x \end{align*}

Thus, the desired integral is obtained by setting t=-\frac{1}{2}. Hence,

    \begin{align*} \frac{1}{4}\int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x &= \frac{1}{4} \cdot \Gamma' \left ( 1-\frac{1}{2} \right )\\ &= \frac{1}{4} \cdot \Gamma' \left ( \frac{1}{2} \right )\\ &=\frac{1}{4} \cdot \psi^{(0)} \left ( \frac{1}{2} \right ) \Gamma \left ( \frac{1}{2} \right ) \\ &= \frac{1}{4} \cdot \left ( -2\ln 2 -\gamma \right ) \cdot \sqrt{\pi} \\ &= -\frac{\sqrt{\pi}}{4} \left ( 2\ln 2+\gamma \right ) \end{align*}

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Series of zeta sum

Let \zeta denote the Riemann zeta function. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left( n - \sum_{k=2}^{n} \zeta(k) \right)\]

Solution

Let a \in \mathbb{R} \mid \left| a \right|<2. We are proving the more general result.

    \[\sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) = a \left ( \frac{\psi^{(0)}\left ( 2-a \right ) + \gamma}{1-a} + 1 \right )\]

where \psi^{(0)} denotes the digamma function.

First of all, we note that:

    \begin{align*} \sum_{k=2}^{n} \zeta(k) &= \sum_{k=2}^{n} \sum_{m=1}^{\infty} \frac{1}{m^k} \\ &= \sum_{m=1}^{\infty} \sum_{k=2}^{n} \frac{1}{m^k}\\ &=\sum_{m=1}^{\infty} \sum_{k=0}^{n-2} \frac{1}{m^{k+2}} \\ &=\sum_{m=1}^{\infty} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} \\ &=\sum_{m=1}^{1} \frac{1}{m^2} \sum_{k=0}^{n-2} \frac{1}{m^k} + \sum_{m=2}^{\infty} \frac{1}{m^2}\sum_{k=0}^{n-2} \frac{1}{m^k} \\ &= n-1 + \sum_{m=2}^{\infty} \frac{1}{m^2} \left ( \frac{1-\frac{1}{m^{n-1}}}{1-\frac{1}{m}} \right ) \\ &= n-1 +\sum_{m=2}^{\infty} \frac{m^{n-1}-1}{\left ( m-1 \right ) m^n} \\ & = n -1 + \sum_{m=1}^{\infty} \frac{1}{m(m+1)} - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \\ &= n - \sum_{m=2}^{\infty} \frac{1}{(m-1) m^n} \end{align*}

Thus,

    \begin{align*} \sum_{n=1}^{\infty} a^n \left ( n - \sum_{k=2}^{n} \zeta(k) \right ) &= \sum_{m=2}^{\infty} \frac{1}{m-1}\sum_{n=2}^{\infty} \left ( \frac{a}{m} \right )^n \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1) m \left ( m-a \right )} \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{m-a} \left ( \frac{1}{m-1} - \frac{1}{m} \right ) \\ &=a^2 \sum_{m=2}^{\infty} \frac{1}{(m-1)(m-a)} -a^2 \sum_{m=2}^{\infty} \frac{1}{m(m-a)} \\ &=a^2 \sum_{m=1}^{\infty} \frac{1}{m(m+1-a)} - a^2 \sum_{m=1}^{\infty} \frac{1}{m(m-a)} + \\ & \quad \quad \quad \quad \quad + \frac{a^2}{1-a} \\ &=\frac{a^2}{1-a} \left ( \psi^{(0)} (2-a) + \gamma \right ) +\\ & \quad \quad \quad \quad \quad + a \left ( \psi^{(0)} (1-a) + \gamma \right )+\frac{a^2}{1-a}\\ &= a \left ( \frac{\psi^{(0)}(2-a) + \gamma}{1-a} + 1 \right ) \end{align*}

due to the reflection formula \displaystyle \psi^{(0)} (z+1) = \psi^{(0)}(z) + \frac{1}{z}.

Side note: If a=1 then the sum equals \zeta(2) - 1.

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