# Does the Weierstrass M-test show analyticity?

I'm trying to show (textbook exercise) that the riemann-zeta function is analytic. The solution is here:

Why does the proof say that the zeta series converges to an analytic function? Doesn't the M-test merely show uniform convergence? The zeta series (whose term is inside the first modulus in the above solution) isn't a power series, so I can't argue that convergence implies analyticity either.

• What book is that?
– lhf
May 22, 2013 at 13:07
• Complex Analysis by Saff & Snider. May 22, 2013 at 17:16

The limit of a sequence of analytic functions that converges uniformly is an analytic function.

This can be proved by combining uniform convergence with Morera's theorem because uniform convergence allows you to switch limit and integral.

http://en.wikipedia.org/wiki/Morera%27s_theorem#Uniform_limits

It is hard to read the mind of the book author or to see an error in the question, but for the same reason that zeta converges absolutely for $\Re s > 1$, its derivatives converge absolutely. By uniformity of convergence, whatever the limit of zeta is, it is complex differentiable in $\Re s > 1$ (because summation and differentiation operations commute under uniform convergence), and therefore analytic.