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(1) Show that the Riemann \( \zeta \) function defined by \[ \displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \] is analytic on the set \( \displaystyle A = \left\{ s \in \mathbb{C} | Re(z) > 1 \right\} \) and write a convergent series for \( \zeta^{\prime}(s) \) on that set.


(2) Show that, while the Riemann \( \zeta \) function is uniformly convergent on every closed disk contained in A, it is not uniformly convergent on all of A.


(3) Show that \[ \displaystyle \sum_{n=1}^{\infty} \frac{ [ \log n ]^{k} }{n^{s}} \] is analytic on the set A for every \( \displaystyle k \in \mathbb{N} \).

Hello Nickos,

i'll answer the first one for now.

It is known that \( \zeta (s) \) converges absolutely because the integral test is applied. (I'll add a proof later).
Now I'll show that \(\zeta \) is analytic on the half plane. Indeed using Morena's Theorem it is sufficient to show that \( \displaystyle \oint_\gamma\zeta (z)\,dz=\oint _\gamma \sum_{n\geq 1}\frac{1}{n^z}\,dz \quad (1) \) equals zero for all closed contours on the half plane.

Because \( \gamma \) is a closed contour on the open half plane , there exists \( \epsilon>0 \) such that \( \mathfrak{Re}(z)>1+\epsilon \) for all \( z \in \gamma \) . Hence, \( \displaystyle \left | \frac{1}{n^z} \right |\leq \frac{1}{n^{1+\epsilon}} \) forall \( z \in \gamma \).

Since \( \displaystyle \sum_{n\geq 1} \frac{1}{n^{1+\epsilon}} \) converges uniformly (proof omitted) we can alter summation and integration back at \( (1) \) thus we get \( \displaystyle \sum_{n\geq 1}\oint _\gamma \frac{1}{n^z}\,dz \). But each integral is zero, by Cauchy's Theorem \((*) \) , thus we get our desired result.

\((*) \) Note that \( \displaystyle \frac{1}{n^z}=e^{-z\ln z} \) which is clearly analytic in \(z \) for \(n \geq 1 \).

The derivative has a formula of \( \displaystyle \zeta'(s)=-\sum_{n\geq 2} \frac{\ln n}{n^s} \) (after we justify the proof of differentiation) which can be obtained as:
$$\begin{aligned}
\frac{d\zeta}{dz} &=\frac{d}{dz}\left ( \sum_{n\geq 1} n^{-z} \right ) \\
&= \sum_{n\geq 1}\frac{d}{dz}n^{-z}\\
&= -\sum_{n\geq 1} \left ( \ln n \cdot n^{-z} \right )\\
&= -\sum_{n\geq 2}\frac{\ln n}{n^z}
\end{aligned}$$

I owe a proof here , that I add it today... Sorry for not coming back sooner.

The Riemann \( \mathbf{\zeta} \) function converges absolutely.
Indeed let \( z=x+yi \) be a complex number. Since the sum \( \displaystyle \sum_{n\geq 1} \frac{1}{n^x},\; x>1 \) converges (by the integral test) we see that: $$\left | \zeta\left ( z \right ) \right |\leq \sum_{n\geq 1}\frac{1}{n^x}\leq 1+\int_{1}^\infty\frac{da}{a^x}=1+\frac{1}{1-x}$$ Thus \(\zeta \) converges absolutely and the proof is over.

For the sake of completeness we begin by stating the Analytic Convergence theorem, which will be used throughout the solution.

Analytic Convergence Theorem : Let \( \displaystyle A \) be an open subset of \( \displaystyle \mathbb{C} \) and let \( \displaystyle \left( g_{k} \right)_{k\in\mathbb{N}} \) be a sequence of analytic functions defined on \( \displaystyle A \). If \( \displaystyle g(z) = \sum_{k=1}^{\infty} g_{k}(z) \) converges uniformly on every closed disk contained in \( \displaystyle A \), then \( \displaystyle g \) is analytic on \( \displaystyle A \, , \, g^{\prime}(z) = \sum_{k=1}^{\infty} g^{\prime}_{k}(z) \) pointwise on \( \displaystyle A \) and also uniformly on every closed disk contained in \( \displaystyle A\).

Now we are ready to begin with the proofs.

(1) Let \( \displaystyle B \) be a closed disk in \( \displaystyle A \) and let \( \displaystyle \delta \) be its distance from the line \( \displaystyle Re(z) = 1 \). We have that \[ \displaystyle |n^{-s}| = |{e}^{-s\log n}| = {e}^{-Re(s)\log n} = n^{-Re(s)} \,.\] If \( \displaystyle s \in B \), then \( \displaystyle Re(s) \geq 1 + \delta \), so \( \displaystyle |n^{-s}| \leq n^{-(1+\delta)} , \forall s \in B \). Set \( \displaystyle M_{n} = n^{-(1+\delta)} , n \in \mathbb{N} \), and observe that \( \displaystyle \sum_{n=1}^{\infty} M_{n}
\) converges. So, by the Weierstrass M-Test we have that the riemann \( \displaystyle \zeta \) function converges uniformly on \( \displaystyle B \).

From the Analytic Convergence Theorem follows that the riemann \( \displaystyle \zeta \) function is analytic on \( \displaystyle A \) and that we can differentiate term by term to get \[ \displaystyle \zeta^{\prime}(s) = - \sum_{k=1}^{\infty} ( \log n ) n^{-s} = - \sum_{k=2}^{\infty} \frac{ \log n }{ n^{s} }, \] which converges on \( \displaystyle A \) and also uniformly on every closed disk contained in \( \displaystyle A\).


(2) Suppose that the riemann \( \displaystyle \zeta \) function converges uniformly on (all of) \( \displaystyle A \). Then,by Cauchy's criterion, for \( \displaystyle \epsilon = \frac{1}{8} \) there exists \( \displaystyle n_{0} \in \mathbb{N} \) such that
\[ \Big| \sum_{n=m_{1}}^{m_{2}} \frac{1}{n^{s}} \Big| < \frac{1}{2} , \; \forall m_{2} \geq m_{1} \geq n_{0} , \forall s \in A \] Therefore, for \( \displaystyle s = 1 + \frac{1}{k} \in A \), where \( \displaystyle k \in \mathbb{N} \), we have that
\begin{align*}
\frac{1}{8} &>
\frac{1}{m_{1}^{1+\frac{1}{k}}} + \frac{1}{( m_{1}+1)^{1+\frac{1}{k}}} + \dots + \frac{1}{( m_{1}+m_{2})^{1+\frac{1}{k}}} \\
&\geq \frac{ m_{2} + 1 }{ (m_{1} + m_{2})^{1+\frac{1}{k}}} \\
&\geq \frac{ m_{2} + 1 }{ (2m_{2})^{1+\frac{1}{k}}} \\
&> \frac{ m_{2} }{ (2m_{2})^{1+\frac{1}{k}}} \\
&= \frac{1}{2} \cdot \frac{1}{ (2m_{2})^{\frac{1}{k}}}
\end{align*}
If \( \displaystyle k \in \mathbb{N} \) is chosen large and such that \[ \displaystyle \Big| \frac{1}{(2m_{2})^{\frac{1}{k}}} - 1 \Big| < \frac{1}{2} , \] then the above hold for that \( \displaystyle k \) as well and, additionally, we have that \( \displaystyle \frac{1}{2} \cdot \frac{1}{ (2m_{2})^{\frac{1}{k}}} > \frac{1}{4} \). We have now reached a contradiction, since we have that \[ \frac{1}{8} > \frac{1}{2} \cdot \frac{1}{ (2m_{2})^{\frac{1}{k}}} > \frac{1}{4} \]Hence the riemann \( \displaystyle \zeta \) function does not converge uniformly on \( \displaystyle A \).


(3) Repeated use of the Analytic Convergence Theorem on the derivatives of the riemann \( \displaystyle \zeta \) function gives the desired result.