I need some feedback on the following:
According to the page about the von Mangoldt function at the Mathworld page, the Riemann hypothesis is equivalent to the statement:
$$\psi = x + \mathcal{O}(\sqrt{x} \log ^2(x))$$
where $\psi$ is the summatory von Mangoldt function.
$\psi(x) -x$ can be calculated as the following limit:
$$\psi(x) - x = \lim_{s\to 1} \, \zeta (s) \sum _{n=2}^x \left(\sum _{d|n} \frac{\mu (\text{d}(n))}{\text{d}(n)^{s-1}}\right)-x$$
where $\zeta (s)$ is the Riemann zeta function $\mu$ is the Möbius function and the $\sum\limits_{d|n}$ is the sum over the divisors $d$ that divide $n$.
The term $\sqrt{x} \log ^2(x)$ inside the big Oh-notation parentheses $\mathcal{O}$ can be calculated as the following limit:
$$\sqrt{x} \log ^2(x) = \lim_{s\to 1} \, \frac{\left(1-\frac{1}{x^{s-1}}\right)^2 \zeta (s)}{\frac{1-\frac{1}{\exp ^{s^{\frac{1}{\sqrt{x}}}-1}(x)}}{x}}$$
Putting these two expressions together we have:
$$\psi - x = \mathcal{O}(\sqrt{x} \log ^2(x))$$
$$\lim_{s\to 1} \, \zeta (s) \sum _{n=2}^x \left(\sum _{d|n} \frac{\mu (\text{d}(n))}{\text{d}(n)^{s-1}}\right)-x = \mathcal{O}\left(\lim_{s\to 1} \, \frac{\left(1-\frac{1}{x^{s-1}}\right)^2 \zeta (s)}{\frac{1-\frac{1}{\exp ^{s^{\frac{1}{\sqrt{x}}}-1}(x)}}{x}}\right)$$
Then I would divide by $\zeta (s)$ but I don't know if it is allowed when it is inside a limit. Please tell me if I am wrong. However dividing by $\zeta (s)$ we get:
$$\lim_{s\to 1} \, \sum _{n=2}^x \left(\sum _{d|n} \frac{\mu (\text{d}(n))}{\text{d}(n)^{s-1}}\right)-\frac{x}{\zeta (s)} = \mathcal{O}\left(\lim_{s\to 1} \, \frac{\left(1-\frac{1}{x^{s-1}}\right)^2}{\frac{1-\frac{1}{\exp ^{s^{\frac{1}{\sqrt{x}}}-1}(x)}}{x}}\right)$$
From here on I don't know how to evaluate the truth or falsity of this so I resort to the very crude non-scientific method of using a computer program like Mathematica:
Clear[n, s, nn]
nn = 12
Print["Limit[Table[Zeta[s]*(1-1/n^(s-1))^2/((1-1/Exp[n]^((s^(n^(-1/2))\
-1)))/n),{n,1,nn}],s\[Rule]1]"]
Limit[Table[
Zeta[s]*(1 - 1/n^(s - 1))^2/((1 - 1/Exp[n]^((s^(n^(-1/2)) - 1)))/
n), {n, 2, nn}], s -> 1]
Print["Limit[Table[-k+Zeta[s]*(Total[Table[Total[1/Divisors[n]^(s-1)*\
MoebiusMu[Divisors[n]]],{n,1,k}]]),{k,1,nn}],s\[Rule]1]"]
Limit[Table[-k +
Zeta[s]*(Sum[
Total[1/Divisors[n]^(s - 1)*MoebiusMu[Divisors[n]]], {n, 2,
k}]), {k, 2, nn}], s -> 1]
s = 9999999999999999/10000000000000000
a = Table[-k/
Zeta[s] + (Total[
Table[Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)], {n, 2,
k}]]), {k, 2, nn}];
Print["Left hand side"]
N[a, 10]
b = Table[(1 - 1/n^(s - 1))^2/((1 - 1/Exp[n]^((s^(n^(-1/2)) - 1)))/
n), {n, 2, nn}];
Print["Right hand side"]
N[b, 10]
For $s = 9999999999999999/10000000000000000$ and for $x \geq 2$
I get the output
During evaluation of In[257]:= Left hand side
Out[266]= {-6.794631684*10^-17, -2.090496922*10^-16, \
-3.843624111*10^-16, -5.792065402*10^-16, -7.863846758*10^-16, \
-1.001831277*10^-15, -1.223033703*10^-15, -1.448338753*10^-15, \
-1.676607395*10^-15, -1.907026665*10^-15, -2.138999975*10^-15}
During evaluation of In[257]:= Right hand side
Out[269]= {1.306852819*10^-16, 1.208240531*10^-16, 1.515093350*10^-16,
9.056554378*10^-17, 1.905655438*10^-16, 9.597452887*10^-17,
1.266598108*10^-16, 1.167985819*10^-16, 2.167985819*10^-16,
7.700905467*10^-17, 1.770090547*10^-16}
I notice that the values are close to zero. Also I might have mixed $n$ and and $x$ in the formulas on the right hand side when I wrote the latex. The question is if the equation in the latex before the computer program where I divided by the zeta function, is correct.
The limit of $s$ going towards 1 gives zero on both sides:
Clear[s]
a = Table[
Limit[-k/
Zeta[s] + (Total[
Table[Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)], {n, 2,
k}]]), s -> 1], {k, 2, nn}];
Print["Left hand side"]
N[a, 10]
b = Table[
Limit[(1 - 1/n^(s - 1))^2/((1 - 1/Exp[n]^((s^(n^(-1/2)) - 1)))/n),
s -> 1], {n, 2, nn}];
Print["Right hand side"]
N[b, 10]
For $s$ approching $1$ and $x \geq 2$:
During evaluation of In[323]:= Left hand side
Out[333]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
During evaluation of In[323]:= Right hand side
Out[336]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}