Joy of Mathematics

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,

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

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

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

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.

Limit with zeta and Gamma function

August 13, 2018 / Leave a comment

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.

On a log Gamma integral using Riemann sums

August 9, 2018 / Leave a comment

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:

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