Wednesday, December 31, 2014

Divergence of sum of reciprocals of primes

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 $2^{2n+2}$).

Relation between Gamma Function and Zeta Function

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

Monday, December 29, 2014

Arithmetic Geometric Series

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 $\dfrac{a+b+c+d}{4} \ge d \Rightarrow \dfrac{a+b+c}{4}+\dfrac{d}{4} =d \Rightarrow \dfrac{a+b+c}{4} = \dfrac{3d}{4} \Rightarrow \dfrac{a+b+c}{3} \ge d \Rightarrow \dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$

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