# The non-trivial zeroes of the Riemann Zeta Function

Hi I have read extensively about the Riemann zeta function and there is something that I am partly confused about. It is my belief that a positive number raised to any power is also positive since it is just a root and/or power of that same positive number. Using this info, I do not understand how there could be any non-trivial zeroes in the positive numbers because raising 1,2,3,4,5,6,... to any power will be positive and the reciprocal of a positive number is still positive. What my essential question is how does the sum of exponentially smaller and smaller $positive$ numbers end up reaching zero?

• The Riemann zeta function is defined as the analytic continuation of the map $\left\{ \begin{matrix} \mathbb{R}_{> 1} & \to & \mathbb{R} \\ s & \mapsto & \displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^{s}} \end{matrix} \right\}$ to $\mathbb{C} \setminus \{ 1 \}$. It is this analytic continuation that has non-trivial zeros. – Transcendental Sep 9 '16 at 6:20
• this thread and this answer concerning the partial sum may clarify things too. – Raymond Manzoni Sep 9 '16 at 6:36
• non-trivial means you have to read a complex-analysis course prior to $\zeta(s)$ – reuns Sep 9 '16 at 6:57

Your reasoning holds over the real numbers (which we call $\mathbb R$). It fails over the complex numbers.

Consider $3^s$ where $s$ is a complex number. In fact, in honor of the Riemann zeta function, let $s = \frac{1}{2}+bi$, where $b$ is real (this is precisely the critical line). Then, we have that: $$3^s = 3^{(1/2)+ib} = 3^{1/2}+3^{ib}=\sqrt{3}\times 3^{ib}$$ How do we make sense of $3^{ib}$? The normal way is to notice that $3^{ib} = e^{(\ln 3)ib}$ Finally, we have an identity: $$e^{ix} = \cos x+i\sin x$$ This gives us that: $$3^s = \sqrt{3}(\cos(\ln(3)b)+i\sin(\ln(3)b)$$ Now, this is negative for infinitely many $b$. One obvious" choice is to first write that $\ln(3)b = x$, then notice that $$3^s = \sqrt{3}(\cos x+i\sin x)$$ Is negative for $x = \pi+2\pi k$ (among other numbers). This corresponds with $b = (1/\ln(3))(\pi+2\pi k)$. We have that: $$3^{(1/2)+i\pi/\ln(3)} = \sqrt{3}(-1+0) = -\sqrt{3}$$

The question does not depend on summing the series, it can be asked for the partial sums. There are complex zeros of any of the partial sums, for example, $1 + 2^{-s} = 0$ when $s$ is an odd integer multiple of $\frac{i \pi}{\log 2}$.

This is a consequence of any exponential function $a^s$ with $a$ real and positive, extending to a function of all complex $s$, but the extension can assume any nonzero complex value.

• (you should better talk of the partial sums of $\eta(s)$, whose zeros converge to the zeros of $\zeta(s)$ or $1-2^{1-s}$) – reuns Sep 9 '16 at 6:56
• @RaymondManzoni no, the zeros of the partial sums $\sum_{n=1}^N n^{-s}$ don't converge to the zeros of $\zeta(s)$, since for $Re(s) \in (0,1)$ : $\zeta(s) = \frac{1}{s-1} + \sum_{n=1}^\infty (n^{-s} - \int_n^{n+1} x^{-s}dx) = \lim_{N \to \infty} \frac{N^{1-s}}{s-1} +\sum_{n=1}^N n^{-s}$ – reuns Sep 9 '16 at 7:58
• @RaymondManzoni so you should look instead at the zeros of $\frac{N^{1-s}}{s-1} + \sum_{n=1}^N n^{-s}$ – reuns Sep 9 '16 at 7:59
• @RaymondManzoni so you don't want to understand that adding $\frac{(N+1)^{1-s}}{s-1}$ to $\sum_{n=1}^N n^{-s}$ reduces drastically the error (and hence improves the localization of the zeros) – reuns Sep 9 '16 at 9:01
• @RaymondManzoni yes, because I'm sure you didn't get it : for $Re(s) > 0$, $\zeta(s) = \frac{(N+1)^{1-s}}{s-1}+ (\sum_{n=1}^N n^{-s}) + \mathcal{O}(2N^{-Re(s)})$ where the error term is really bounded by $2 N^{-Re(s)}$ independently of $Im(s)$. Hence, when $N$ isn't very small (independently of $Im(s)$) you can't decently imagine that any zero of $\zeta(s)$ is close to a zero of $\sum_{n=1}^N n^{-s}$ – reuns Sep 9 '16 at 9:33