Lets define the following function series (Riemann Zeta Function):  $$\varsigma (z)=\sum _{n=1}^\infty n^{-z} $$  where the power is defined by the primary branch of the Logarithm.  A) prove that the series converges in : $ D= \left\{ z:\operatorname{Re}(z) &gt; 1 \right\}$ and also converges absolutely and uniformly in $D_\delta = \left\{z:\operatorname{Re}(z)&gt;1+ \delta \right\}  $  B) find $\varsigma '(z) $.  Do guys have any hints? I thought about calculating the radius of convergence, but how do I do it in this case? and how do I prvoe that it converges with this minus power? also I didnt get the information about the power being defined that way, im not sure also that I translated this accurately from hebrew.
I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example  and .  But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)&gt; 1$.  I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!  On $\Re(s)=\sigma&gt; 1$, we have  $$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$  and  $$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$  for any $\epsilon &gt;0$, so by the direct comparison test, if   $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$  converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.  By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral  $$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$  which holds for any $\epsilon &lt; \infty$.  Any correction or improvement is welcomed! Thanks in advance.
describes the first-order vs. second-order logic as follows:     First-order logic uses only variables that range over individuals (elements of the domain of discourse); second-order logic has these variables as well as additional variables that range over sets of individuals.   It gives $\forall P\,\forall x (x \in P \lor x \notin P)$ as an SO-logic formula, which makes perfect sense to me.  However, in a , the poster claimed that $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$ is an FO formula. I think this must not be the case, since in the above formula, $x$ and $y$ are sets of individuals, while $z$ is an individual (and therefore this must be an SO formula).  I mentioned this as a comment, but two users commented that ZF can be entirely described by FO logic, and therefore $\forall x\forall y(x=y\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))$ is an FO formula.  I'm confused. Could someone explain this please? 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An example is in order here:  static string NumberToWords(int number) {     string[] numbers = new string[]          { "", "one", "two", "three", "four", "five",            "six", "seven", "eight", "nine" };     string[] tens = new string[]          { "", "", "twenty", "thirty", "forty", "fifty",            "sixty", "seventy", "eighty", "ninety" };     string[] teens = new string[]         { "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen",           "sixteen", "seventeen", "eighteen", "nineteen" };      string ans = "";     switch (number.ToString().Length)     {         case 3:             ans += string.Format("{0} hundred and ", numbers[number / 100]);         case 2:             int t = (number / 10) % 10;             if (t == 1)             {                 ans += teens[number % 10];                 break;             }             else if (t &gt; 1)                 ans += string.Format("{0}-", tens[t]);         case 1:             int o = number % 10;             ans += numbers[o];              break;         default:             throw new ArgumentException("number");     }     return ans; }   The smart people are cringing because the string[]s should be declared outside the function: well, they are, this is just an example.  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When I post a question and want to add a new line, I have to press Return two times, or else the next text will appear on the same line.  Is this really supposed to behave this way? If so, why? Changing the style of the first *typeset* line of a paragraph Assume $0 &lt; a_0 \leq a_1 \leq \cdots \leq a_n$, show that equation   $$ p_n(z) = a_0 z^n + a_1 z^{n-1} + \cdots + a_{n-1} z + a_n = 0 $$   does not have root in $|z| &lt; 1$. Here $z$ is complex.   I started by considering the case in which $n = 2$, and tried to use the triangle inequality, while no progress was made. Any hints will be appreciated. When and how did "momentarily" come to mean "in a moment", rather than "for a moment"? "Momentarily" used to mean "for a moment" only, and not "in a moment". Thus, newscasters could be divided into two clear groups: those who would say "we'll be back momentarily," and those who would not.  This restriction made sense to me, because having both definitions would promote ambiguity if a unique interpretation could not always be derived from the context. But in recent years it seems "momentarily" is regularly, maybe even more often, used to mean "in a moment" by newscasters of every caliber, and in fact this is even shown to be the definition when looked up in most dictionaries.  When did this word's meaning change? How did it come about? Is lemonade better when made with simple syrup? On a hot summer day and with a basketful of lemons I often want a nice cold glass of lemonade. A lot of recipes suggest making lemonade with simple syrup, which requires cooking, whereas other recipes just have stirred-in sugar.  What exactly are the benefits of making lemonade (or limeade) with simple syrup versus just stirring in sugar? "May I go to the bathroom?" and "I asked if I might go to the bathroom."   The modal auxiliary 'may/might' is used to ask permission.     "He may not have understood your question." and "He might not have understood your question."   The same modal auxiliary is used for uncertainty.  Why? Prove $p_n(x) \rightarrow \sqrt{x}$ uniformly as $n \rightarrow \infty$ Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_{n+1}(x) = p_n(x) + \frac{1}{2}(x - p_n(x)^2)$.  Prove that $p_n(x) \rightarrow \sqrt{x}$ uniformly for $0 \leq x \leq 1$ as $n \rightarrow \infty$.  I'm not sure where to go with this.  I can show inductively that $\forall n,\  0 \leq p_n(x) \leq 1$ for $0 \leq x \leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me.  I know that I need to show that $\forall \epsilon&gt;0: \exists N \in \mathbb{N}: \forall n \geq N: \forall x\in [0,1]: \big|p_n(x) - \sqrt{x}\big| &lt; \epsilon$ Edit:  I just had a thought.  I'm given a version of Dini's theorem that reads  Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that: (i) $f_n \rightarrow f$ pointwise on $X$, and (ii) $f_n(x) \geq f_{n+1}(x) $ for all $x \in X, n \in \mathbb{N}$. Then: $f_n \rightarrow f$ uniformly on $X$.  Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - \sqrt{x}$?  Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued.  The only part I would need to prove is (i), which is where I get confused.  Correct me if I'm wrong, but after that I could say that since $(-p_n)_n \rightarrow -\sqrt{x}$, then $(p_n)_n \rightarrow \sqrt{x}$.  Any advice?