Simplest, Quickest and Self-Contained Proof that1/2≤θ≤1(θ Being the Least Upper Bound of the Real Parts of the Zeros of ζ(s))

We will prove that ζ(s) (s=σ+it), defined by
ζ(s)=Σn^-s (σ>1), (1)
=(n^-s-(n+u)^-s)du+1/s-1 (σ>0),(2)
=(1-p^-s)^-1(σ>1),(1)
has infinity of complex zeros in a σ≥1/2-δ for every constant δ>0. (In (3) the product runs over all primes p). Let θ denote the least upper bound of the real parts of the zeros of ζ(s). Then by (3) we have trivially θ≤1 and by what we will prove it follows that θ≥1/2. These results are not new. But the merit of our proof is the fact that apart from using Cauchy’s Theorem for certain rectangles, we use only the simple facts given by (1), (2) and (3). The proof is simpler than the one given in [1]. Without complicating the proof we prove the following theorem.