Series of zeta sum

October 20, 2018 / Leave a comment

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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Lebesgue measure of Cantor set

October 2, 2018 / Leave a comment

An alternative way to define the Cantor set is the following:

    \[\mathfrak{C}=[0,1] \setminus \bigcup_{n=1}^\infty \;\bigcup_{k=0}^{3^{n-1}-1} \left(\frac{3k+1}{3^n},\frac{3k+2}{3^n}\right)\]

What is the Lebesgue measure of the Cantor set if we consider it as a subset of \mathbb{R}? Is \mathfrak{C} countable?

A divergent series …. or maybe not?

September 23, 2018 / Leave a comment

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

September 13, 2018 / Leave a comment

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

August 25, 2018 / Leave a comment

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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