[Part II] The first attempt in the difference of pi(x) and Li(x)
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.
\[[\pi(k)-\text{Li}(k)]\geq\frac{\alpha k}{\ln^n k}\geq\frac{\alpha \xi}{\ln^n \xi}\]
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.
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.
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