We know that the Riemann Zeta function $$\sum_{n=1}^{\infty} \frac{1}{n^s}$$ converges when $Re(s) > 1$. How do you prove the convergence of the series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{(z+n)^s}$$ converges when $Re(s) > 0$ and $z\in \mathbb{C}$ except for some $z$. Any idea what restriction we need to have on $z$ besides $z$ cannot be negative integer?
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1$\begingroup$ This is the Hurwitz zeta function, and proofs of its absolutel convergence in $\Re s>1$ for fixed $z$ are essentially the same as for the zeta function itself (compare the tail of the series to a $p$-integral). $\endgroup$– Greg MartinCommented Jun 17, 2020 at 4:52
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1$\begingroup$ @Greg Martin Can I ask you another question? What if the series is alternating? How do I prove that? $\endgroup$– Jk_0fjnewuifCommented Jun 17, 2020 at 4:54
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1$\begingroup$ @Smith: For absolute convergence, it doesn't matter if the series is alternating or not. $\endgroup$– metamorphyCommented Jun 17, 2020 at 6:09
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$\begingroup$ The (first-year calculus) alternating series test shows that it converges when $s$ and $z$ are real and $s>0$. And then there are ways (standard in the study of Dirichlet series) to deduce that if the series converges for some $s_0\in\Bbb C$, then it converges for every $s\in\Bbb C$ with $\Re s>\Re s_0$. You can also pair the terms manually and prove the absolute convergence of the resulting series, perhaps using the mean value theorem to estimate the paired terms. $\endgroup$– Greg MartinCommented Jun 17, 2020 at 7:00
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$\begingroup$ @GregMartin Is there any reasons when Re(q)>0 for the Hurwiz zeta functions, $q$ is the notation in the link that you share. If Re(q)<0, does that affects the convergence of that series? $\endgroup$– Jk_0fjnewuifCommented Jun 17, 2020 at 13:54
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You asked the same question there
If $z$ is a negative integer then $(z+n)^{-s}$ is undefined.
Thus we assume it is not. For $\Re(s) > 1$ it converges absolutely. For $\Re(s) > 0$ we need a partial summation
$$\sum_{n=1}^N (-1)^n (z+n)^{-s} = (\sum_{n=1}^N (-1)^n) (z+N)^{-s}+\sum_{n=1}^{N-1}(\sum_{m=1}^n (-1)^m) ((z+m)^{-s}-(z+m+1)^{-s})$$ The first term $\to 0$, and for the second term we use that $$(z+m)^{-s}-(z+m+1)^{-s} = \int_0^1 s(z+x)^{-s-1}dx= O(s(z+m)^{-s-1})$$ Finally we let $N\to \infty$.
