# Harmonic series principal value

Harmonic series $$1+1/2+1/3+1/4+\cdots$$ is divergent.

However if we take the generalized Hurwitz harmonic series

$$F(s)=\sum_{n=0}^{\infty}(n+a)^{-1+s}+\sum_{n=0}^{\infty}(n+a)^{-1-s}$$

can we say or regularize the result so $F(s) \to -2\Psi (a)$ , when $s \to 0$ ??

For example, near $0$ the Function $F(s)$ has the expansion $$F(s)= -\Psi (a)+ \frac{1}{s},$$ so the limit $F(0+\epsilon)+F(0-\epsilon )$ should be finite.

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An 'intuitive' fast answer : the sum is $\zeta(1)$ with a single pole there from the Stieltjes expansion at $z=1$ : $\zeta(z)=\frac 1{z-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}\gamma_n\;(z-1)^n$ so that you'll have to remove this pole to get Euler's constant $\gamma$. –  Raymond Manzoni Jul 2 '13 at 12:30
$\zeta (1+s)= \frac{1}{s}+ \gamma$ , $\zeta (1-s)= \frac{1}{-s}+ \gamma$ so the sum gives $\zeta (1+s)+ \zeta (1-s)= 2\gamma$ in the limit $s \to 0$ –  Jose Garcia Jul 2 '13 at 12:45
Yes @Jose this should be right even if $\zeta$ without the simple pole is not really zeta anymore (your solution is in fact different since you remove half of the powers of $s$). Cheers, –  Raymond Manzoni Jul 2 '13 at 12:54
What is the question ?. –  Felix Marin Jul 29 '14 at 0:14