I apologize for not having posted in roughly a week, but this past week has been a bit hectic. The week before spring break always seems to be. Anyway, I thought I would move us a bit further along in our goal of proving the Prime Number Theorem. In the post PNT I, we showed that the zeta function has no zeroes for , and in this post, we will show that has no zeroes for . The proof, which I believe was first given by Vallee-Poussin, is rather elegant.

We first need an inequality for , which is a consequence of an elementary trigonometric identity.

**Lemma 1**

*Proof:*Recall the identity . Using the Euler product for , we have

We see that . Hence,

which is since the argument is .

**Theorem 2** * for . *

*Proof:*Suppose for some with . Since is analytic at , we can expand in a power series about in some open disk . So for ,

where we use the fact that a power series converges absolutely strictly within its radius of convergence. Recall that from the meromorphic extension of to the half plane that we obtain the estimate as . Since is analytic at and therefore is bounded as , we obtain

which yields a contradiction for sufficiently small.

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