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For $0 < |z| < 1$ and $\Re(s) > 1$ $$F(s,z) = \Gamma(s)\sum_{n=1}^\infty z^n n^{-s} =\int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{-nx} z^n dx= \int_0^\infty \frac{x^{s-1}}{e^x/z-1}dx$$ $$= \frac{ x^s …

$f:M_k(\Bbb{C})\to M_k(\Bbb{C})$ is analytic on $\{ A\in M_k(\Bbb{C}),\sigma(A)<1\}$ and not beyond, where $\sigma(A)=\sup |\lambda_j(A)|$ is the spectral radius. If $\sigma(A)< 1$ then write the J …

The binomial series gives a meromorphic continuation of any series of the form $\sum_{n=N}^\infty g(n) f(n)^{-s}$ for some polynomials $g(x) = \sum_{j=0}^t c_j x^j$ and $f(x) = x^r \prod_{k=1}^d (x-a_ …

I abandoned making a precise answer but for a continuous branch of $(N+1/2 +ix)^{-s}$, $F_N(s)= \int_{-\infty}^{\infty} \frac{(N+1/2 +ix)^{-s}}{\cosh(\pi x)}\,dx$ converges and is analytic for $\Im(s …

Summing by parts, using the binomial series and inverting the double sum works fine, obtaining $$\zeta(s,a)-s \zeta(s,N) = \sum_{n=1}^{N-1} n ((n+a-1)^{-s}-(n+a)^{-s})\\+ \sum_{k=0}^\infty {-s \cho …

Your question doesn't make much sense. Let $a_d =\sum_{m|d} \mu(m) \frac{e^{-(d/m)/2}}{d/m}=O(1)$. As $\frac{\zeta'(s)}{\zeta(s)}=O(2^{-s})$ then for $\Re(s) >0$, $F(s) = \sum_{d\ge 1} a_d\frac{\zeta' …