How to prove the following limit is converging and is equal to $0$ Because when I tried to sketch the graph on desmos while changing the variable a I don't seem to find it converging ?
$$\lim_{n\to \infty}\Sigma_{r= 1}^{n} \frac{1}{\sqrt{r}}\cos(\ln(\frac{1}{r^a}))$$,where a $\in$ ${\rm I\!R}$ - $\{0\}$
How I obtained this question:
I did a little simplification to Zeta function, in the following way,
$\zeta(s) = \Sigma_{r=1}^{\infty}(\frac{1}{r^s} = r^{-s})$
I set s = $\frac{1}{2} + a \iota$ , a $\in$ ${\rm I\!R}$ - $\{0\}$
Now, $r^{-s} = r^{-\frac{1}{2} - a\iota} = r^{-\frac{1}{2}}\times r^{-a\iota} = \frac{1}{\sqrt{r}}e^{\ln(r^{-a})\cdot\iota} = \frac{1}{\sqrt{r}}\cdot(\cos(\ln(\frac{1}{r^{a}}))+(\sin(\ln(\frac{1}{r^{a}})\iota)))$
Now, Wikipedia states that
Thus, if the hypothesis is correct, all the non-trivial zeros lie on the critical line consisting of the complex numbers 1 / 2 + i t, where t is a real number and i is the imaginary unit
Someone on mathoverflow stated
the Riemann hypothesis is verified for the first 10 trillion zeroes
How did someone verify that?(I tried using cmath and mpmath together in python and it outputs a non-zero complex number for various values of s in $\zeta(s)$)
If a complex number z = $x+\iota y = 0$ then $x = 0$ and $y = 0$ provided $x,y \in $ ${\rm I\!R}$
So, considering only the real part of my simplification, I want to know how the summation tends to $0$(without the summation it is pretty clear that the limit tends to $0$ as $\frac{1}{\sqrt{r}} \to 0$).
In sum, I want to know,
(1) How it is verified for 1st 10 trillion zeroes
(2) How to prove the above limit as converging
Any help is appreciated.
It diverges because for $\Re(s) > 0$ (taking the real part of the following with $s=1/2+i a$) $$\sum_{n=1}^N n^{-s}-\frac{(N+1)^{1-s}-1}{1-s}=\sum_{n=1}^N \int_n^{n+1}(n^{-s}-x^{-s})dx=\sum_{n=1}^N \int_n^{n+1}\int_n^x st^{-s-1}dtdx$$ which in absolute value is $$ \le \sum_{n=1}^N \int_n^{n+1} |s| x^{-\Re(s)-1}dx= \int_1^{N+1} |s| x^{-\Re(s)-1}dx=|s|\frac{(N+1)^{-\Re(s)}-1}{-\Re(s)}$$ The same argument shows that for $\Re(s) > 0$ $$\zeta(s)-\frac1{s-1}=\lim_{N\to \infty} (\sum_{n=1}^N n^{-s}-\frac{(N+1)^{1-s}-1}{1-s})$$
Also, again, the zeros of $\Re(\zeta(s))$ are not isolated, they are a bunch of curves traversing the critical strip, you can't locate the non-trivial zeros just from it.