So I was looking at the Riemann hypothesis and I saw the relationship between the Riemann zeta function and the Dirichlet eta function which really confused me because I didn't understand how a divergent series could somehow become convergent simply by multiplying it by a constant (for 0 < s < 1). So I tried to derive the relationship myself and encountered something that I thought might explain the issue.
So, after factoring out the (2^(1-s)), the infinite series that is being multiplied by it only goes to (n/2), not n. Now, I understand how this won't matter for (s > 1), because all the terms at infinity will not add up to anything since the series is convergent for that interval. However, for (0 < s < 1), I'm unsure about why the two series should be considered equal. I know that at infinity you are generally supposed to ignore the difference between (n) and (n/2), but in this case I think it is important.
The reason being that if you for example subtract the series that goes to (n/2) from the normal Riemann zeta function, you will get (n/2) extra terms past the ((n/2)th) term. In this case however, the extra (infinity/2) terms will keep adding up to (infinity) because the series diverges for (0 < s < 1). So I don't think the two series should be considered equal
I was wondering if someone could help clear up this confusion, I also understand that it might be some kind of analytic continuation so it doesn't have to be exactly correct?
Thanks.
Best Answer
The important thing is that $\zeta(s) = \sum n^{-s}$ only for $\Re(s) > 1$. One can the prove that with that definition of the zeta function, for $\Re(s) > 1$, we have
$$\zeta (s) = (1-2^{1-s})^{-1}\eta(s)$$
However, the right-hand side is defined for $\Re(s) > 0$ (with a simple pole at $s = 1$), so we can "extend" the zeta function to obtain a new function which coincides with our original function in its domain of definition.
By the identity theorem, there is a unique way of extending our original function. This process is called analytic continuation.