The non-trivial zeroes of the Riemann Zeta Function

Question

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?

Answer 1

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: $$3^{(1/2)+i\pi/\ln(3)} = \sqrt{3}(-1+0) = -\sqrt{3}$$

Answer 2

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.