### Delhi Government gives paid holiday for its employees and students to learn Vipassana

Labels: aam admi party, AAP, Arvind Kejriwal, Ashok, Buddhism, Goenka, Kiran bedi, Manish sisodia, Vipassana

This is a short piece to summarize whats going in my life.

This would put Kejriwal and Manish Sisodia the practicing duo into the league of Ashoka. Vipassana is now going to be taught in schools and the teachers can get a paid leave to learn this technique. This is perhaps the biggest social engineering step that AAP has taken and its good step in the right direction. This is a step none of the opposition can publicly oppose especially Kiran Bedi who brought this to Tihar Jail and got lot of publicity. Going by her tweets its very clear that she doesn't practice or even understand Vipassana but will be tongue tied to speak against it. The biggest political beneficiary will be AAP and biggest loosers will be current political parties. which is great for the country. Another great step by AAP.

Labels: aam admi party, AAP, Arvind Kejriwal, Ashok, Buddhism, Goenka, Kiran bedi, Manish sisodia, Vipassana

We know that Harmonic series $\sum \frac{1}{n} $ is diverging. The proof is very simple by grouping the terms. Whereas for geometric sequence $\frac{1}{2^n}$ is converging. Our question is what happens when we have $\sum_{p \in prime} \frac{1}{p}$.

We will prove this by contradiction. Assuming that the sum of reciprocals of prime $S_p$ is a constant value. As $\frac{1}{p_i}$ are positive constants. That means for some $\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_{n-1}} < S_n -\frac{1}{2}$ and

$\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_n}+\frac{1}{p_{n}} \ge S_n - \frac{1}{2}$.

Thus $\frac{1}{p_{n+1}}+\frac{1}{p_{n+2}}+ ... < \frac{1}{2}$ and multiplying both sides by positive $x$ we get $\frac{x}{p_{n+1}}+\frac{x}{p_{n+2}}+ ... < \frac{x}{2}$

To prove this we first define a function $N(x)$ which counts the number of numbers whose prime factor is among the first $ k$ primes. For example if $k = 4$ the primes are $\{ 2,3,5,7 \}$. So the function $N(10) = 10, N(15) = 13, N(27) = 20$. ie for $10$ all the factors are from prime numbers $\{ 2,3,5,7 \}$. Let $k$ be the number of first k primes and $x$ be any number and we try to find $N(x)$. We see that any number $y \le x$ can be written as $y = {p_1}^{a_1}{p_2}^{a_2}...{p_t}^{a_t}$. We have to count all those $y$ whose factors are among the first $k$ primes. We realise that any number can be written as $uv$ where $w^2v$ or $u = w^2$ is a square number and so $v$ is a product of primes less than $k^{th}$ prime and has the form $p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}...p_k^{\alpha_k}$ where $\alpha_1,\alpha_2 ...\alpha_k \in \{0,1\}$. Thus $N(x) \le \sqrt{x}2^{k}$ and the numbers which are not divisible by first k primes are $x-N(x) \le \frac{x}{p_{k+1}}+\frac{x}{p_{k+2}}+\frac{x}{p_{k+3}}+....$.

So there are two equations as below

$\frac{x}{p_{n+1}}+\frac{x}{p_{n+2}}+ ... < \frac{x}{2}$

$x-N(x) \le \frac{x}{p_{k+1}}+\frac{x}{p_{k+2}}+\frac{x}{p_{k+3}}+....$.

Combining we have $x-N(x) < \frac{x}{2} \Rightarrow \frac{x}{2} < N(x)$ and holds for all values of $x$. Let $k = n $ then $N(x) \le 2^n\sqrt{x}$. Therefore we havce $\frac{x}{2} < N(x) \le 2^n\sqrt{x}$. When $n = 2^{2n+2}$ we get $\frac{1}{2}2^{2n+1} < N(x) \le 2^{2n+1}$. This contradiction proves that the series P diverges. ( A similar contradiction is obtained for $x$ equal to any value greater than $2^{2n+2}$).

We know that $\Gamma(x) = \int_0^\infty e^{-t}t^{x-1} dt$

Let $t = ru \implies dt = r du$ Which means

$\Gamma(x) = \int_0^\infty e^{-ru} {(ru)}^{x-1} r du$

$\Gamma (x) = \int_0^\infty r^x u^{x-1} e^{-ru} du$

$\frac{1}{r^x} \Gamma (x) = \int_0^\infty u^{x-1} e^{-ru} du$

Taking sum on both the sides we get

$\sum_{r=0}^\infty \frac{1}{r^x} \Gamma (x) = \sum_{r=0}^{\infty}\int_0^\infty r^x u^{x-1} e^{-ru} du$

$\Gamma(x) \sum_{r = 0}^{\infty} \frac{1}{r^x} = \int_0^{\infty}u^{x-1} \sum_{r=0}^{\infty}e^{-ru} du$

$\Gamma(x) \zeta(x) = \int_{0}^{\infty}u^{x-1}\frac{e^{-u}}{1-e^{-u}} du$

This is a famous inequality that one should know. The case for two numbers is really easy $(\sqrt{a}-\sqrt{b})^2 \ge 0 \Rightarrow a+b \ge 2\sqrt{a}\sqrt{b} \Rightarrow \frac{a+b}{2} \ge \sqrt{ab}$

The case for 4 variables is also pretty straight forward $\dfrac{a+b+c+d}{4} =\dfrac{ \dfrac{a+b}{2} +\dfrac{c+d}{2}}{2} \ge \dfrac{\sqrt{ab}+\sqrt{cd}}{2} \ge \sqrt{ \sqrt{ab} \sqrt{cd}} \ge \sqrt[4]{abcd}$

The case for 3 variables is a little bit tricky but can be proved quite easily if we use use the 4 variable proof. Let $d = \sqrt[3]{abc}$. Then $\dfrac{a+b+c+d}{4} \ge \sqrt[4]{abcd} =\sqrt[4]{abc(abc)^{\frac{1}{3}}} = \sqrt[4]{(abc)^{\frac{4}{3}}} =(abc)^{\frac{1}{3}} =d$ and so we have $\dfrac{a+b+c+d}{4} \ge d \Rightarrow \dfrac{a+b+c}{4}+\dfrac{d}{4} =d \Rightarrow \dfrac{a+b+c}{4} = \dfrac{3d}{4} \Rightarrow \dfrac{a+b+c}{3} \ge d \Rightarrow \dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$

A man is like fraction. The numerator corresponds to who he is and the denominator corresponds to what he honks of himself. He greater the numerator, smaller is the fraction.