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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

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