Trivial zeros of the Riemann zeta function

[mrbohn1]

Clearly I am missing something obvious here, but how is it that negative even numbers are zeros of the Riemann zeta function?

For example:

$$\zeta (-2)=1+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+...=1+4+9+..$$

Which is clearly not zero. What is it that I am doing wrong?

 

[Santa1]

You are using the definition of zeta(s) for Re(s) > 1 with a number that has a real part smaller than or equal to 1.

 

[CRGreathouse]

What is it that I am doing wrong?

You need not the function you posted, but its analytic continuation.

 

[mrbohn1]

Thanks! It all becomes clear.

 

[blue_raver22]

The Riemann zeta function is a mathematical function that is widely used in number theory and has many interesting properties. One of these properties is that it has infinitely many zeros, known as the trivial zeros, which lie on the negative even numbers on the complex plane.

It is important to note that the Riemann zeta function is defined for complex numbers, not just real numbers. This means that the input can be a combination of both real and imaginary numbers. When we plug in a negative even number into the zeta function, we are actually plugging in a complex number with an imaginary component of zero. This is because the negative even numbers can be written as -2n, where n is a positive integer. Therefore, when we plug in -2n into the zeta function, we are actually calculating \zeta (-2n) = \sum_{k=1}^\infty k^{-2n} which simplifies to \sum_{k=1}^\infty \frac{1}{k^{2n}}.

Now, let's look at what happens when n = 1. In this case, we have \zeta (-2) = \sum_{k=1}^\infty \frac{1}{k^{2}} = 1+\frac{1}{4}+\frac{1}{9}+... which is the same series that you have written in your question. However, this series does not converge to zero, but rather to a finite value of approximately 1.645. This is because the Riemann zeta function has a special property called analytic continuation, which allows it to be extended to values outside of its original domain. In this case, the zeta function is extended to the negative even numbers on the complex plane, and the series converges to a finite value at these points.

Therefore, it is not that you are doing something wrong, but rather that the Riemann zeta function has unique and interesting properties that may seem counterintuitive at first. I hope this explanation helps to clarify why the negative even numbers are considered to be trivial zeros of the Riemann zeta function.