Simpler zeta zeros

Question

Is it true that

$$\lim_{y\rightarrow\infty}\dfrac{\sum_{n=1}^{y}n^{-1/2-iy}}{\zeta(1/2+iy)}=1$$

? Below is a plot of $$\sum_{n=1}^{y}\dfrac{1}{n^{s}}\text{for }s=\dfrac{1}{2}+iy$$

set against its smooth analytic continuation. Is this less expensive computationally for large $y$?

Notes

x = 1/2; Plot[{Re[\!\( \*SubsuperscriptBox[\(\[Sum]\), \(n = 1\), \(y\)] 
\*FractionBox[\(1\), SuperscriptBox[\(n\), \(x + I\ y\)]]\)], 
Re[Zeta[x + I y]], Im[\!\( \*SubsuperscriptBox[\(\[Sum]\), \(n = 1\), \(y\)] 
\*FractionBox[\(1\), SuperscriptBox[\(n\), \(x + I\ y\)]]\)], 
Im[Zeta[x + I y]]}, {y, 0, 30}] 

Included with subsuperscript boxes for ease of reading when pasted into Mathematica.

Update

To address Mercio's point:

The plot of the quotients (as posed in the original question) is very messy (includes grid-line at $1$), and clearly doesn't "tend to $1$" in the strict sense of the meaning. Perhaps if zeros of each function were excepted, it would make more sense. Suggestions of re-phrasings of the question are welcomed!

Update 2

Question now rephrased - improvements suggested by Raymond Manzoni:

Is it true that $$\lim_{y\rightarrow\infty}\zeta(\dfrac{1}{2}+iy)+\dfrac {1} {2}\dfrac {1} {[y/\pi]^{1/2 + iy}} - \sum_ {n = 1}^{[y/ \pi]}\dfrac {1} {n^{1/2 + iy}}=0?$$

which John M has more or less answered below.

Answer 1

(Updated)
As a complement to John M's proof (+1) I'll add some 'visual clarifications' (whatever this may mean...) about the behavior of the finite sum of $\zeta\;$ for a large but fixed ordinate $y\in\mathbb{R}^+$ : $$\tag{1}S_y(N)=\sum_{k=1}^N\frac 1{k^{\,1/2+iy}}$$ Let's suppose that $y$ is the first ordinate of a nontrivial zero larger than $10000$ and represent all the partial sums $\,S_y(N)$ for $\;N=1\cdots 3183=[y/\pi]\;$ in the complex plane :

The cross at the middle is at the origin ($z=0$) while '1' represents $z=S_y(1)=1$ and '2' $\;S_y(2)=1+\dfrac {2^{-iy}}{\sqrt{2}}$ and so on up to '3183' representing $\;S_y(3183)$ very near to $0$ again. $$-$$ The picture shows figures looking like Cornu (or Euler) spirals. Let's justify this :
the sum $S_y(N)$ is obtained by addition of terms $\,\displaystyle \frac 1{k^{1/2+iy}}=\frac {e^{-iy\log(k)}}{\sqrt{k}}$.
The next term will thus be $\,\displaystyle \frac {e^{-iy\log(k+1)}}{\sqrt{k+1}}$.
Now for $k\gg 1$ the denominator will change only slightly while the change of phase of the numerator will be $\;\delta=-y\;(\log(k+1)-\log(k))=-y\,\log(1+1/k)\approx -\dfrac yk$.

$\delta\approx -\dfrac {y}k$ gives a special role to the values of $k$ such that $\dfrac yk\approx f\pi$ with $f$ integer :

• for $f=2m$ (i.e. $k=\left[ \frac y{2m\pi}\right]$) we have $\,\delta\approx -2m\pi\,$ : for values of $k$ near of $\left[ \frac y{2m\pi}\right]$ the terms have nearly the same phase and their addition will nearly give a straight line (see f=2, f=4, and so on on the picture)
• for $f=2m+1$ (i.e. $k=\left[ \frac y{(2m+1)\pi}\right]$) we have $\,\delta\approx -(2m+1)\pi\,$ : for values of $k$ near of $\left[ \frac y{(2m+1)\pi}\right]$ two consecutive terms nearly cancel each other and that's what is happening in the middle of the 'nodes' f=1, f=3, f=5 and so on.

Let's zoom the center of the final node $f=1$ :

The line nearly crossing the origin is obtained with the term $k=3183=\left[ y/\pi\right]$ while the line at its left and right came from $k=3182$ and $k=3181$ respectively. The value $S_y(3183)$ itself is far out of the picture (say $12$ or more times higher) while $S_y(3182)$ is far at the bottom on the other side. Taking the middle of these two values should bring us not too far from our target even if the different partial sums $S_y(N)$ don't really 'go down to $0$' ! (they merely turn around it)

All this explains (but doesn't prove) that an excellent approximation for $\zeta$ may be obtained (near the zeros at least) with the formula : $$\zeta\left(\frac 12+iy\right)\approx -\frac 1{2\,[y/\pi]^{1/2+iy}}+\sum_{n=1}^{[y/ \pi]}\frac{1}{n^{1/2+iy}}$$

(numerically the absolute error appears majored by $\dfrac{2.5}{y^{3/2}}$ in the range $(10,10000)$, and probably for larger values, while the initial sum from the question was majored by $\dfrac{0.9}{y^{1/2}}$)

Now what happens when we continue adding terms after $\dfrac y{\pi}$ ? Well the difference of phase $\,\delta\,$ will become smaller than $\pi$ and we will turn round and round and obtain an ever growing ball of black wool :-) (illustration for $N=10^7$)

This discussion was rather qualitative and may be followed by these more general and precise expansions provided not only for $\,N=\left[\dfrac y{\pi}\right]\,$ terms but also for $\left[\dfrac y{e}\right]$ terms with $0<e<2\pi$.

But the story doesn't end here and Riemann himself found that you didn't need to compute the sum of $[y]$ terms (or $\left[\dfrac y{\pi}\right]$ or $\left[\dfrac y{2\pi}\right]$ or whatever) but that $\left[\sqrt{\dfrac y{2\pi}}\right]$ terms were enough, at least if you accept some correction terms! (Riemann-Siegel formula)
and basic facts about Riemann $\zeta$ and the Euler Maclaurin formula.

As a fun alternative you may examine with care my first picture : notice that the distance from f=1 to f=3 is $1$, that the distance from f=3 to f=5 is $\dfrac 1{\sqrt{2}}$ and think 'Symmetry'!

All the pictures presented here were produced interactively using the (old) CSE applet that I created for Matthew R. Watkins (see the comments in case of problems).
Vitaliy Kaurov kindly provided nicer pictures using Mathematica.

Answer 2

Please allow me to write what is true:

We have the estimate in the critical strip ($s = \sigma + it$): $$\zeta(s) = \sum_{n < N} n^{-s} + \frac{N^{1-s}}{s-1} + O(N^{-\sigma}).$$ Therefore, $$\zeta\left(\frac{1}{2}+it\right) = \sum_{n < t} \frac{1}{n^{\frac{1}{2}+it}} + O(t^{-1/2}).$$ So $$\lim_{t \rightarrow \infty} \left | \zeta\left(\frac{1}{2}+it\right) - \sum_{n < t} \frac{1}{n^{\frac{1}{2}+it}} \right| = 0.$$

You may also be interested in this recent preprint.

ADDED LATER (at Daniel's request):

First, we assume that $\sigma = \operatorname{Re}{z} > 1$, so that the sum $\sum n^{-s}$ converges absolutely. We can write the sum $\sum_{n \geq N} n^{-s}$ in terms as a Stieltjes integral, $$\sum_{n = N}^\infty \frac{1}{n^s} = N^{-s} + \int_N^\infty x^{-s} \;d(\lfloor x \rfloor),$$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. Now we can integrate by parts to get $$\int_N^\infty x^{-s} \;d(\lfloor x \rfloor) = -N^{-s+1} + s \int_N^\infty x^{-s-1} \lfloor x \rfloor \; dx.$$ Let $\{x\}$ denote the fractional part of $x$, i.e. $\{x\} = x - \lfloor x \rfloor$. We get $$\zeta(s) = \sum_{n < N} n^{-s} + N^{-s} + \frac{N^{1-s}}{s-1} - s \int_N^\infty x^{-s-1} \{x\} \;dx.$$ Since $\{x\} < 1$, the integral on the right actually converges for all $\sigma = \operatorname{Re}(s) > 0$, i.e. we get an analytic continuation of the Riemann zeta function from $\sigma > 1$ to $\sigma > 0$. In fact, for $\sigma>0$, we can bound the integral, $$\left| \int_N^\infty x^{-s-1} \{x\} \;dx \right| < \frac{N^{-\sigma}}{\sigma}.$$

Anyway, this is all standard material available in any book on analytic number theory.