Why is $\zeta(s)=\lim_{x\to\infty}\left(\sum_{n\leqslant x} \frac{1}{n^s}-\frac{x^{1-s}}{1-s}\right)$ for $0

Question

I am currently reading the book Introduction to Algebraic Number Theory by Apostol. To introduce some important asymptotic formulas, Apostol gives a rough definition of the Riemann zeta function (for $s\in\mathbb{R}^+$),

$$\begin{equation}\zeta(s)=\begin{cases} \sum_{n}n^{-s}, &s>1\\ \lim_{x\to\infty}\left(\sum_{n\leqslant x} \frac{1}{n^s}-\frac{x^{1-s}}{1-s}\right), &0<s<1 \end{cases}\end{equation}$$

The second part really confused me. How could we approach this limit? If we see $\zeta$ as an analytic continuation of $\sum_{n}\frac{1}{n^s}$, it should be written as $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1} dx$$ This formula can be easily derived from $\Gamma(s)=\int_0^\infty x^{s-1}e^{-x}dx$ by substituting $x=nu$ (which was exactly what Riemann did in his paper). However, I don't see the connection between this formula and the limit form for $0<s<1$. I am really new to this function so maybe this is a dumb question. But please point it out why we can write $\zeta(s)$ in the limit form for real $0<s<1$.

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Also, historically, is the limit form derived from the formal or the converse?

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Thanks in advance, any help will be appreciated.

Answer 1

For an elementary approach, you want to show that the limit is indeed analytic in $s$ (uniform limit of analytic functions) in an open subset of $D=\{s:\Re(s)>0\land s\ne1\}$. For $\Re(s)>1$ this is fairly trivial since $x^{1-s}\to0$. For $0<\Re(s)\le1$, a full asymptotic expansion makes this more obvious, but it suffices to simply bound the error between the given sum and

$$\int_0^x\frac{\mathrm dt}{t^s}=\frac{x^{1-s}}{1-s}\tag{$0<\Re(s)\le1,s\ne1$}$$

using something such as Taylor expansions.

A much more general approach is given by the Euler-Maclaurin summation formula, which states that

$$\sum_{n\le x}\frac1{n^s}=\zeta(s)+\frac1{(1-s)x^{s-1}}+\frac1{2x^s}-\frac s{12x^{s+1}}+\mathcal O(x^{-s-3})$$

For $\Re(s)>1$, every term after $\zeta(s)$ tends to zero, so we get

$$\lim_{x\to\infty}\sum_{n\le x}\frac1{n^s}=\zeta(s)$$

For $\Re(s)>0$, the $x^{-s+1}$ term needn't go to zero, so we get

$$\lim_{x\to\infty}\left[\sum_{n\le x}\frac1{n^s}-\frac1{(1-s)x^{s-1}}\right]=\zeta(s)$$

In general, by moving all terms which don't go to zero to the other side, we may get a converging limit expression for $\zeta(s)$ for $\Re(s)>-N$ for any natural $N$. It is interesting to note that this gives exacts when $s$ is a negative integer since $\sum_{n\le x}n^{-s}$ has a closed form.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\lim_{x \to \infty}\pars{\sum_{n\ \leqslant\ x}{1 \over n^{s}} - {x^{1 - s} \over 1 - s}}\right\vert_{\ 0\ <\ s\ <\ 1} = \lim_{x \to \infty}\bracks{\zeta\pars{s} + s\int_{x}^{\infty}{\braces{t} \over t^{s + 1}}\,\dd t} \end{align} where I used a Zeta Function Identity. However, $\ds{0 < \verts{s\int_{x}^{\infty}{\braces{t} \over t^{s + 1}} \,\dd t} < {1 \over x^{s}}}$. Then, $$ \bbx{\left.\lim_{x \to \infty}\pars{\sum_{n\ \leqslant\ x}{1 \over n^{s}} - {x^{1 - s} \over 1 - s}}\right\vert_{\ 0\ <\ s\ <\ 1} = \zeta\pars{s}}\\ $$