A divergent series …. or maybe not?

The number n ranges over all possible powers with both the base and the exponent positive integers greater than n, assuming each such value only once. Prove that:

    \[\sum_{n} \frac{1}{n-1}=1\]

Let us denote by \mathcal{M} the set of positive integers greater than 1 that are not perfect powers ( i.e are not of the form a^p , where a is a positive integer and p \geq 2 ).  Since the terms of the series are positive , we can freely permute them. Thus,

    \begin{align*} \sum_{n} \frac{1}{n-1} &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \frac{1}{m^k-1} \\ &= \sum_{m \in \mathcal{M}} \sum_{k=2}^{\infty} \sum_{j=1}^{\infty} \frac{1}{m^{kj}}\\ &=\sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{kj}} \\ &= \sum_{m \in \mathcal{M}} \sum_{j=1}^{\infty} \frac{1}{m^j \left ( m^j-1 \right )} \\ &= \sum_{n=2}^{\infty} \frac{1}{n\left ( n-1 \right )} \\ &= \sum_{n=2}^{\infty} \left ( \frac{1}{n-1} - \frac{1}{n} \right )\\ &= 1 \end{align*}

 

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Inequality for a Lipschitz function

Let f:[0, 1] \rightarrow \mathbb{R} be a Lipschitz function . Prove that forall n \in \mathbb{N} it holds that

    \[\left|\int_{0}^{1} f(x) \, \mathrm{d}x -\frac{1}{n} \sum_{k=1}^{n} f \left ( \frac{k}{n} \right) \right |\leq \frac{M}{2n}\]

Solution

We have that

    \begin{align*} \left|\int_{0}^{1}f(x)\; \mathrm{d} x-\frac{1}{n}\sum_{k=1}^{n}f \left ( \frac{k}{n} \right )\right|&=\left|\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n} \bigg(f(x)-f\left ( \frac{k}{n} \right ) \bigg) \, \mathrm{d}x \right|\\ &\leq M\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}\left|x- \frac{k}{n} \right| \; \mathrm{d} x\\ &=M\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}\left(\frac{k}{n}-x \right)dx\\ &=M\left(-\int_{0}^{1}x \; \mathrm{d}x+\frac{1}{n^{2}}\sum_{k=1}^{n}k\right)\\ &=M\left(-\frac{1}{2}+\dfrac{1}{n^{2}}\cdot\dfrac{n^{2}+n}{2}\right)\\ &=\frac{M}{2n} \end{align*}

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Iterated exponential integral

Prove that

    \[\int_0^{2\pi} \exp \left ( \exp \left ( \exp (it) \right ) \right )\, \mathrm{d}t = 2\pi e\]

Solution

We begin by stating a lemma:

Lemma: Let f be an analytic function on some closed disk  \mathbb{D} which has center a and radius r. Let \mathcal{C} denote the the boundary of the disk. It holds that

    \[\int_{0}^{2\pi} f \left ( a + re^{i \theta} \right )\, \mathrm{d}\theta = 2 \pi f(a)\]

Proof: By the Cauchy integral formula we have that

    \[f(a) = \frac{1}{2\pi i }\oint \limits_{\mathcal{C}} \frac{f(z)}{z-a} \, \mathrm{d}z\]

The equation of a circle of radius r and centre a is given by z=a + re^{i \theta}. Hence,

    \begin{align*} f(a) &= \frac{1}{2\pi}\int_{0}^{2\pi}\frac{f(a + re^{i\theta})ire^{i\theta}}{re^{i\theta}}\,\mathrm{d}\theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}f(a+re^{i\theta})\,\mathrm{d}\theta \end{align*}

and the proof is complete.

Something quickie: Given the assumptions in Gauss’ MVT, we have

    \[\left|f(a)|\right\leq \frac{1}{2\pi}\int_{0}^{2\pi}\left|f(a+re^{i\theta})\right|\,\mathrm{d} \theta\]

The proof of the result is pretty straight forward by using the fact that

    \[\left | \int_{a}^{b} f(x) \, \mathrm{d}x \right | \leq \int_{a}^{b} \left|f(x) \right|  \, \mathrm{d}x\]

Back to the problem the result now follows by the lemma.

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Limit with zeta and Gamma function

Let \zeta denote the Riemann zeta function and \Gamma denote the Euler’s Gamma function. Prove that

    \[\frac{1}{2} \lim_{n \rightarrow +\infty} \left [ \zeta\left ( 1 + \frac{1}{n} \right ) - \Gamma \left ( \frac{1}{n} \right ) \right ] = \gamma\]

where \gamma stands for the Euler – Mascheroni constant.

Solution

Using the Laurent expansion of the \zeta function we have that

    \begin{align*} \zeta(s) \sim \frac{1}{s-1} + \gamma + \mathcal{O} \left ( s-1 \right ) &\Rightarrow \zeta\left ( 1 + \frac{1}{n} \right ) \sim \frac{1}{1+\frac{1}{n} - 1} + \gamma + \\ & \quad \quad \quad \quad \quad \quad \quad +\mathcal{O} \left ( 1 + \frac{1}{n} - 1 \right )\\ &\Rightarrow \zeta \left ( 1 + \frac{1}{n} \right ) \sim n + \gamma + \mathcal{O} \left ( \frac{1}{n} \right) \end{align*}

Using the Laurent expansion of the \Gamma function we have that

    \begin{align*} \Gamma(s) \sim \frac{1}{s} - \gamma + \mathcal{O}(s) &\Rightarrow \Gamma \left ( \frac{1}{n} \right ) \sim n - \gamma + \mathcal{O} \left ( \frac{1}{n} \right ) \end{align*}

The result now follows.

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On a log Gamma integral using Riemann sums

Evaluate the integral

    \[\mathcal{J} = \int_0^1 \log \Gamma (x) \, \mathrm{d}x\]

using Riemann sums.

Solution

Partition the interval [0, 1] into n subintervals of length \frac{1}{n}. This produces,

(1)   \begin{equation*}  \int_{0}^{1} \log \Gamma(x) \, \mathrm{d}x = \lim_{n \rightarrow +\infty} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) \end{equation*}

On the other hand, assuming n is even:

    \begin{align*} \frac{1}{n} \sum_{k=1}^{n} \log \Gamma \left ( \frac{k}{n} \right ) &= \frac{1}{n} \log \prod_{k=1}^{n} \Gamma \left ( \frac{k}{n} \right ) \\ &= \frac{1}{n} \log \prod_{k=1}^{n/2} \Gamma \left ( \frac{k}{n} \right ) \Gamma \left ( 1 - \frac{k}{n} \right )\\ &=\frac{1}{n} \log \prod_{k=1}^{n/2} \frac{\pi}{\sin \frac{\pi k}{n}} \\ &= \log \sqrt{\pi} - \log \left ( \prod_{k=1}^{n} \sin \frac{\pi k}{n} \right )^{1/n}\\ &= \log \frac{\sqrt{2 \pi}}{\left ( 2n \right )^{1/2n}} \end{align*}

since it holds that

    \[\prod_{k=1}^{n} \sin \frac{\pi k}{n} = \frac{n}{2^{n-1}}\]

Euler’s Gamma reflection formula was used at line (3). Letting n \rightarrow +\infty we get that

    \[\int_0^1 \log \Gamma(x) \, \mathrm{d}x = \log \sqrt{2\pi}\]

If n is odd we work similarly.

 

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