The zeta Function, The Line of Conditional Convergence

The Line of Conditional Convergence

If a series is convergent its terms go to zero, which means they are all less than 1 beyond some point. Divide the n^th term by n^1+ε, and the resulting series is absolutely convergent by the integral test. If you're dealing with a series of complex numbers, they are eventually restricted to the interior of the unit circle, and when divided by n^1+ε, their norms form an absolutely convergent series, hence the modified series of complex numbers is absolutely convergent.

Returning to the world of zeta, ζ(f,s) may be well defined for values of s to the left of s_a, but the series is conditionally convergent on these points. If s is increased by anything larger than 1, the terms are divided by something larger than n, and the series becomes absolutely convergent. In other words, moving s to the right one unit runs into or crosses s_a. The conditionally convergent points of zeta are trapped between s_a-1 and s_a.

Let s_c be the greatest lower bound of the real components of s for which ζ(f,s) is defined. In other words, s_c is the x coordinate of the leftmost point of ζ(f,s), or the limit of these points. Draw a vertical line through s_c and call it the line of conditional convergence. Zeta is defined on part of the strip between s_c and s_a, and on everything to the right of s_a.

If ε is a positive real number, convergence at s implies convergence at s+ε. The terms of the convergent series are multiplied by a real (geometric) sequence that approaches 0, whence abel summation implies convergence. The zeta function is well defined on a half line, whose left endpoint lies somewhere between s_c and s_a.

In fact, zeta is defined on the entire half plane to the right of s_c. This isn't hard to prove, if you use the complex variation of abel summation.

Convergence at s implies the partial sums are bounded. Move to the right and up in the complex plane. Call this difference vector v. Build a new sequence 1/n^v. This is multiplied by ζ(f,s), term by term, to get ζ(f,s+v). The real component of v causes the terms of 1/n^v to approach 0. The imaginary component of v makes the terms spin around the origin. However, the angle is multiplied by log(n), so the spinning slows way down. After a while, 1/n^v - 1/(n+1)^v always lies in the same quadrant. (This will be the first or fourth quadrant, depending on whether v moves up or down in the complex plane.) That's all we need for abel summation.

The zeta function converges at s+v, for every vector v that moves up and to the right, or down and to the right. This holds for every s where zeta converges. Therefore zeta converges on the half plane to the right of s_c.

Riemann Zeta

What is s_c for the Riemann zeta function? We know that s_a = 1, and ζ(1) does not converge, so s_c also equals 1. The Riemann zeta function is defined on the half plane to the right of 1.