If $z_n$ are the zeros of the zeta function, what is the limit of $\Im{(z_n)}$ as $n$ goes to infinity? Sorry if this question has already been asked, but it's a little difficult to look things up in Google if the statement of the problem is not very simple and involves symbols that Google doesn't recognize.
The question I have regards the zeta function. If $z_n$ is the sequence of non-trivial zeros of the zeta function with positive imaginary part and sorted by ascending imaginary part, what is the limit when $n$ goes to infinity of $\Im{(z_n)}$?
Does this explode out to infinity or is it finite?
Asking for a friend (paper here).
He has derived a new super simple equation whose solution is equivalent to the Riemann hypothesis.
 A: The Riemann-von Mangoldt formula asserts that the number of zeroes of the form $\frac{1}{2} + it$ where $t \in [0, T]$ is asymptotically
$$\frac{T}{2\pi} \log \frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T)$$
from which it follows that $\text{Im}(z_n)$ grows something like $\frac{2 \pi n}{\log n} \left( 1 + \frac{\log \log n}{\log n} \right)$, but I haven't been too careful about that calculation.
Large tables of zeros are available to double-check this asymptotic against; for example, the millionth zero has imaginary part $\approx 600269$ whereas the asymptotic above gives $\approx 541230$, so it's a bit of an underestimate.
Working a bit more carefully, write $\text{Im}(z_n) = \frac{2 \pi n}{\log n} \left( 1 + e_n \right)$, where $e_n \to 0$ (slowly). Then to match the asymptotic above we need
$$\frac{n}{\log n} (1 + e_n) \log \left( \frac{n}{\log n} (1 + e_n) \right) - \frac{n}{\log n} (1 + e_n) = n + O(\log n).$$
Dividing by $\frac{n}{\log n}$, expanding out, and canceling the dominant term from both sides gives, after some simplification,
$$e_n \log n + (1 + e_n) \log (1 + e_n) - (1 + e_n) \log \log n - (1 + e_n) = O \left( \frac{(\log n)^2}{n} \right).$$
In order for the LHS to have limit $0$ as $n \to \infty$ we see that we need $e_n \approx \frac{\log \log n + 1}{\log n}$. This is already a noticeable improvement; it improves the estimate of the imaginary part of the millionth zero to $\approx 574149$. To do better than this we'll estimate
$$\log (1 + e_n) = e_n + O(e_n^2)$$
(keeping in mind that $O(e_n^2)$ is $O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right)$ which is quite a bit slower than $O \left( \frac{(\log n)^2}{n} \right)$ so this is not best possible), which means the LHS becomes, after some simplification,
$$\left( e_n \log n - \log \log n - 1 \right) - e_n \log \log n + O(e_n^2)$$
so we can improve our estimate some more to $e_n \approx \frac{\log \log n + 1}{\log n - \log \log n}$. This is again a noticeable improvement; now the estimate for the imaginary part of the millionth zero is $\approx 602157$. We have two digits of accuracy now! Altogether, then,
$$\boxed{ \text{Im}(z_n) \approx \frac{2 \pi n}{\log n} \left( 1 + \frac{\log \log n + 1}{\log n - \log \log n} \right) }$$
and with a little more effort one could give a big-$O$ description of the error in this approximation but I'll stop here.
