# 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 $0<s<1$?

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 $$s\in\mathbb{R}^+$$),

$$$$\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

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 $$0. I am really new to this function so maybe this is a dumb question. But please point it out why we can write $$\zeta(s)$$ in the limit form for real $$0.

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

Thanks in advance, any help will be appreciated.

• Show the limit is analytic in $s$ where it exists and that $\lim_{x\to\infty}x^{1-s}/(1-s)=0$ for $\Re(s)>1$. Then it's equivalent to $\sum_{n\ge1}n^{-s}$ for $\Re(s)>1$. – Simply Beautiful Art Apr 4 '20 at 17:56

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.

• Great answer, now I understand why Apostol introduces this after Euler's summation formula. – justadzr Apr 5 '20 at 4:44


• Apostol actually used the definition (as he claims) $$\left.\lim_{x \to \infty}\left(\sum_{n\ \leqslant\ x}{1 \over n^{s}} - {x^{1 - s} \over 1 - s}\right)\right\vert_{\ 0\ <\ s\ <\ 1} = \zeta(s)$$ to derive that identity. – justadzr Aug 3 '20 at 5:01
• @Yourong'DZR'Zang Interesting. I didn’t know that. I’ll check it. Thanks. – Felix Marin Aug 3 '20 at 9:59