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I was taking my Ph.D Analysis qualifying exam and then they asked me to prove the following exercise.

Let $\epsilon > 0$ and $\mathbb{C}_\epsilon := \{z \in \mathbb{C}, \Re(z) > \epsilon \}$.

(a) $$ F(z) = \sum_{n = 1}^{\infty}\frac{1}{n^z}$$ converges uniformly on $\mathbb{C}_{1 + \epsilon}$.

I prove this part easily using the Weierestrass M-Test bounding the terms of this series by a p-series with $p > 1$ and the uniform convergence follows.

(b) Prove that the series $$\sum_{n=1}^{\infty}\int_{n}^{n+1}(\frac{1}{n^z} - \frac{1}{x^z})dx$$ converges uniformly on $\mathbb{C}_{\epsilon}$

(c) Prove that there exists an analytic function $G$ on $\{z \in \mathbb{C}, \Re(z) > 0\}$ such that $$G(z) = F(z) - \frac{1}{z-1}$$ for $\Re(z) > 1$

In part (b) I try to bound the terms as in part (a) but I was not able to show the uniform convergence on $\mathbb{C}_{\epsilon}$. However in my attempt I can show that it was uniformly convergent on $\mathbb{C}_{1+\epsilon}$ but I think it must be a mistake.

For part (c) I have no idea how to approach.

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Partial answer to b). I will prove uniform convergence on bounded subsets of $\mathcal C_\epsilon$. I doubt if the convergence is uniform on the whole of $\mathcal C_\epsilon$. Let $f(x)=x^{-z}$. Then $|f(x)-f(n)| \leq |x-z| |z|n^{-\Re z-1}$ $\leq |z| n^{-\Re z-1}$ for x between n and $n+1$. Hence $|\int _n ^{n+1} |f(x)-f(n)|dx \leq |z| n^{-\Re z-1}$ proving uniform convergence of the series in b) for $|z|$ bounded.

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Proof of c): let $H(z)= \sum _1 ^{\infty} \int _n ^{n+1} x^{-z} dx+\frac 1 {1-z}$. Since $\int _n ^{n+1} x^{-z} dx=\frac {(n+1)^{-z} -n^{-z}} {1-z}$, The N-th partial sum becomes $\frac {(N+1)^{-z} -1} {1-z}$ which converges to $- \frac 1 {1-z}$. It follows that $H(z)=0$ in $\mathcal C_\epsilon$. By b) $F(z)-H(z)- \frac 1 {1-z}$ is analytic in Re (z) >1.

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  • $\begingroup$ Thank you for this. Really appreciated. $\endgroup$ – Richard Clare Jan 13 '18 at 2:20

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