Processing math: 100%

Wednesday, 16 February 2011

Following the post from Part I , where we have


T_1(s)=\zeta(s) - 1 - \frac{1}{s-1}

T_2(s)=\ln(s-1)+\ln \zeta(s),

T_3(s)=\sum_{p \text{ is prime}}\frac{1}{p^s}-\ln\zeta(s).

and T_i(s) has derivative of order n for s\rightarrow 1^+

Now consider 

-\frac{d^n T_3(s)}{d s^n} - \frac{d^n T_2(s)}{ds^n} - \frac{d^{n-1} T_1(s)}{ds^n}

by our previous result, the above derivative exist. So the result is finite.  Looking at canceled terms from -T_1^{(n)}(s) and -T_2^{(n-1)}(s) those are \ln (s-1) and \frac{1}{s-1}, and the canceled term for -T_3^{(n)}(s) and -T_2^{(n)}(s) is \ln \zeta(s). We have the derivative equals

\pm \left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{s}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{s}}\right)

which is finite.  Now we consider the real valued value of s, let say t=\Re \, s. Therefore

\pm\left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{t}}\right)

is finite for t>1, or precisely for t \rightarrow 1^+.

Taking the positive difference we have

S=\left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{t}}\right)

Now consider \pi(x+1)-\pi(x) where \pi(x) is the number of prime not exceeding x. We see that when x is prime then \pi(x+1)-\pi(x) equals to 1, while when x is not prime then \pi(x+1)-\pi(x) equals 0. Thus

\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}=\sum_{k=2}^{\infty}(\pi(k+1)-\pi(k))\frac{\ln^n k}{k^{t}}

Ann expression S can be written as

S=\sum_{k=2}^{\infty}\left[(\pi(k+1)-\pi(k))\frac{\ln^n k}{k^{t}}-\frac{\ln^{n-1} k}{k^{t}}\right]
=\sum_{k=2}^{\infty}\left[\pi(k+1)-\pi(k)-\frac{1}{\ln k}\right]\frac{\ln^n k}{k^{t}}

Now consider

\frac{1}{\ln x} - \int_x^{x+1} \frac{1}{\ln x}\,dx

By the mean value theorem for integral, there exists c_x \in (x,x+1) such that \int_x^{x+1}\frac{1}{\ln x}\,dx=\frac{1}{\ln c_x}. Thus \frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx=\frac{1}{\ln x}-\frac{1}{\ln c_x} for some c_x \in (x,x+1).
Since c_x \rightarrow \infty whenever x \rightarrow \infty, we have

\lim_{x \rightarrow \infty}\frac{\frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx}{\frac{1}{x}}=\lim_{x\rightarrow\infty}\left(\frac{x}{\ln x}-\frac{x}{\ln c_x}\right)=0

Thus there exists K such that for x>K we have \left|\frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx\right|<\frac{1}{x}. We conclude that

T=T^{\prime}+\left|\sum_{k=K}^{\infty}\left(\frac{1}{\ln k}-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^t}\right| < T^{\prime}+\sum_{k=K}^{\infty}\frac{\ln^{n}k}{k^{t}}

which converges to a finite value. Therefore

S+T = \sum_{k=1}^{\infty}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}

converges to a finite value.

Define the offset Logarithmic Integral Function by \text{Li}(x)=\int_2^{x}\frac{1}{\ln x}\, dx. Therefore

\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx=[\pi(k+1)-\text{Li}(k+1)]- [\pi(k)-\text{Li}(k)]\qquad (1)

Now for the theorem itself:

[Theorem] 
For x\geq 2, the inequality |\pi(x)-\text{Li}(x)| < \frac{\alpha x}{\ln^n x} is true for infinitely many x, where \alpha is a positive constant, and for a sufficiently large number n.


To prove this, suppose for the contrary, that is we can find positive constant \alpha, and bound a  number n such that there are only finite value of x such that the inequality is true. Suppose a is the the largest integer value of x, such that the inequality is true, and also satisfies a>e^n, thus for x>a, the inequality is not true and we have

|\text{Li}(x)-\pi(x)|\geq \frac{\alpha x}{\ln^n x}\qquad \qquad \ln x>n

Now let U>a and consider the sum

R=C+\sum_{k=a+1}^{U}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}

where

C=\sum_{k=2}^{a}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}

by equation (1) and then Abel Summation by part we have

\begin{align*}&\sum_{k=a+1}^{U}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}\\ &=\sum_{k=a+1}^{t}\left([\pi(k+1)-\text{Li}(k+1)]- [\pi(k)-\text{Li}(k)]\right)\frac{\ln^{n}k}{k^{t+1}}\\ &=(\pi(U+1)-\text{Li}(U))\frac{\ln^n U}{U^{t}}-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t}\\ & \qquad\qquad-\sum_{k=a+1}^U [\pi(k)-\text{Li}(k)]\left[\frac{\ln^n (k)}{k^t}-\frac{\ln^n (k-1)}{(k-1)^t}\right]\end{align*}


By our hypothesis (\pi(U+1)-\text{Li}(U))\frac{\ln^n U}{U^{s}}>\left(\frac{\alpha U}{\ln^n U}\right)\frac{\ln^n U}{U^{t}}=\frac{\alpha}{U^{s-1}}>0.
Moreover by the mean value theorem, there exists \xi\in[k-1,k] such that

\left[\frac{\ln^n (k)}{k^t}-\frac{\ln^n (k-1)}{(k-1)^s}\right]=\left(\frac{n}{\ln \xi}-t \right)\frac{\ln^{n}\xi}{\xi^{t+1}}

Thus

R>C-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t}+\sum_{k=a+1}^U[\pi(k)-\text{Li}(k)]\left(t-\frac{n}{\ln \xi}\right)\frac{\ln^{n}\xi}{\xi^{t+1}}

Consider f(x)=\frac{\alpha x}{\ln^n x}, then f^{\prime}(x)=\alpha \left(1-\frac{n}{\ln x}\right)\frac{1}{\ln x}>0 since \ln x>n and \alpha>0, thus the function is increasing.

Since for k\geq a+1 and k>\xi we have

[\pi(k)-\text{Li}(k)]\geq\frac{\alpha k}{\ln^n k}\geq\frac{\alpha \xi}{\ln^n \xi}

Writing D=C-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t} we have

R>D+\sum_{k=a+1}^U\left(\frac{\alpha \xi}{\ln^n \xi}\right)\left(s-\frac{n}{\ln \xi}\right)\frac{\ln^{n}\xi}{\xi^{t+1}}=D+\alpha\sum_{k=a+1}^U\left(t-\frac{n}{\ln \xi}\right)\frac{1}{\xi^t}

Also note that t>1 and so t-\frac{n}{\ln \xi} >1-\frac{n}{\ln\xi} thus we have

R>D+\left(1-\frac{n}{\ln\xi}\right)\sum_{k=a+1}^U \frac{1}{\xi^t}.

Letting U \rightarrow \infty and then t\rightarrow 1^+, we have the right hand side is approaching \infty, while we already established that \lim_{t\rightarrow 1^+} R is finite. Contradiction.

Post a Comment: