Dirichlet series
Abscissa of convergence

Suppose

converges for some s 0 ∈ C, ℜ (s 0) > 0. {\displaystyle s_{0}\in \mathbb {C},\Re (s_{0})>0.}

Proof. Note that:

and define

where

By summation by parts we have

Proof. From the definition

so that

which converges as N → ∞ {\displaystyle N\to \infty } whenever ℜ (s) > σ. {\displaystyle \Re (s)>\sigma.} Hence, for every s {\displaystyle s} such that ∑ n = 1 ∞ a n n − s {\textstyle \sum _{n=1}^{\infty }a_{n}n^{-s}} diverges, we have σ ≥ ℜ (s), {\displaystyle \sigma \geq \Re (s),} and this finishes the proof.

Proof. Note that

and A (N) − f (0) → 0 {\displaystyle A(N)-f(0)\to 0} we have by summation by parts, for ℜ (s) > 0 {\displaystyle \Re (s)>0}

Now find N such that for n > N, | A (n) − f (0) | < ε {\displaystyle |A(n)-f(0)|<\varepsilon }

and hence, for every ε > 0 {\displaystyle \varepsilon >0} there is a C {\displaystyle C} such that for σ > 0 {\displaystyle \sigma >0}:
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