I want to show that $$\zeta(z) = \frac{1}{z-1} + \gamma + O(z-1)$$

for $z \rightarrow 1 $. Why is it enough to show this statement for real $z = s \in (1,2)$? We know that $\zeta$ is meromorphic and has only one simple pole at $z=1$ but how does this help?

Now, for $s \in (1,2)$ we can write

$$\zeta(z) - \frac{1}{z-1} = \sum_{n=1}^{\infty} k^{-s} - \int_{1}^{\infty}t^{-s}dt =\sum_{k=1}^{\infty}\int_k^{k+1}(k^{-s} - t^{-s}) dt$$

and since $|k^{-s}-t^{-s}| = |\int_{k}^{t}su^{-s-1}du| \leq |s|k^{-s-1}$ we know that there exists a constant M such that

$$\sum_{k=1}^{\infty}\int_k^{k+1}(k^{-s} - t^{-s}) dt< M$$

for all $s \in (1,2)$. This implies the whole thing is uniform there and that we are allowed pull in limit, in the end we have:

$$\lim_{s \rightarrow 1}(\zeta(z) - \frac{1}{z-1}) = \sum_{k=1}^{\infty} \int_k^{k+1}(k^{-1}-t^{-1}) dt = \sum_{k=1}^{\infty}(k^{-1}-ln(k+1)+ln(k)) = \gamma$$.

I'm still a little confused with big O-notation, how does this imply our statement for general $z$?

  • 3
    $\begingroup$ sigh this is not a duplicate since the Op doesn't want a proof but rather an explanation of the proof already given to them (which the link does not provide) $\endgroup$
    – FShrike
    Jun 22, 2023 at 14:07
  • $\begingroup$ @FShrike Indeed. I misunderstood what was being asked. $\endgroup$ Jun 22, 2023 at 14:14
  • $\begingroup$ @FShrike thank you $\endgroup$
    – hteica
    Jun 22, 2023 at 14:26
  • $\begingroup$ I think you could use the fact that $(z-1)\zeta(z)$ is entire and $(1,2)$ contains an accumulation point to show any power series for $(z-1)\zeta(z)$ that is valid for $(1,2)$ is valid for all $\mathbb C$. $\endgroup$ Jun 22, 2023 at 14:29
  • $\begingroup$ @eyeballfrog but in our case we don't have any power series. We have only shown that a limit of some function converges towards $\gamma$ $\endgroup$
    – hteica
    Jun 22, 2023 at 14:31

1 Answer 1


$\newcommand{\d}{\,\mathrm{d}}$Before I discuss how the claim follows, I disagree with the assertion: "this implies the whole thing is uniform". There are two limit interchanges that need to be made to justify this proof in more detail:

$$\begin{align}\lim_{s\to1^+}\sum_{k=1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t&\overset{?!}{=}\sum_{k=1}^\infty\lim_{s\to1^+}\int_k^{k+1}(k^{-s}-t^{-s})\d t\\&\overset{?!}{=}\sum_{k=1}^\infty\int_k^{k+1}\lim_{s\to1^+}(k^{-s}-t^{-s})\d t\end{align}$$

And the bound: $$\exists M>0,\,\forall1<s<2:\quad\quad\sum_{k=1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t<M$$Is not strong enough to justify either of those limit interchanges. What we need is uniform convergence (if you want to prove it by "making everything uniform") - I need to know that, for all $\epsilon>0$ there is some $N\in\Bbb N$ with $\left|\sum_{k=n+1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t\right|<\epsilon$ for all $1<s<2$ and all $n\ge N$. That has not yet been shown. However, it is really quite easy to fix this using the same ideas: $$\begin{align}\forall n\in\Bbb N,\,\forall1<s<2:\quad\left|\sum_{k=n+1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\d t\right|&\le\sum_{k=n+1}^\infty|s|k^{-s-1}\\&\le\int_n^\infty st^{-s-1}\d t\\&=n^{-s}\\&\le n^{-1}\end{align}$$

So we can be confident the series uniformly converges. That justifies the first limit interchange! We then need to justify: $$\forall k\in\Bbb N:\quad\lim_{s\to1^+}\int_k^{k+1}(k^{-s}-t^{-s})\d t\overset{?!}{=}\int_k^{k+1}\lim_{s\to1^+}(k^{-s}-t^{-s})\d t$$This is also easy to justify by "making everything uniform" (or the dominated convergence theorem...). Note that: $$\begin{align}|k^{-s}-t^{-s}-(k^{-1}-t^{-1})|&=|(t^{-s}-t^{-1})+(k^{-1}-k^{-s})|\\&\le t^{-1}-t^{-s}+k^{-1}-k^{-s}\\&\le2(1-s^{-1})\end{align}$$For all admissible $k,t,s$ (justifying the final inequality with calculus). So the integrand uniformly converges on $[k,k+1]$ to $k^{-1}-t^{-1}$ and we can be happy. Note the convergence is actually uniform on $(1,\infty)$, which is stronger than what we need.

How does this imply our statement for general $z$?

By the machinery of complex analysis. This has already been answered in the comments, I don't have much to add here.

I know $\zeta$, as function holomorphic on $\Bbb C\setminus\{-1\}$, must have a Laurent expansion at $1$. Examining such a Laurent series in light of the fact $(s-1)\zeta(s)$ converges as $s\to 1^+$ on the real axis shows the Laurent series must start with a $(s-1)^{-1}$ coefficient and show $\zeta(s)$ looks like $(s-1)^{-1}+\gamma+a_1(s-1)+a_2(s-2)^2+\cdots=(s-1)^{-1}+\gamma+O(s-1)$. There is simply no other possiblity: the analyticity is highly restrictive.

Note in the case of real analytic functions, we could have something like $(s-1)f(s)=\gamma+\sqrt{s-1}+\cdots$ which is not of the form $\gamma+O(s-1)$... so technically the proof is incomplete; it's the fact we deal with a meromorphic function that forces everything to work out nicely.

  • 1
    $\begingroup$ @hteica I'm making the pedantic point that, what you wrote: $$\sum_{k=1}^\infty\int_k^{k+1}(k^{-s}-t^{-s})\,\mathrm{d}t<M$$By itself is not sufficient. If you had written: $$\sum_{k=1}^\infty\sup_{1\le s\le 2}\int_k^{k+1}(k^{-s}-t^{-s})\,\mathrm{d}t<M$$Then that would have been sufficient (to exchange the first limit, at least...) and it's an easy fix anyway: my justification for exchanging the first limit is using the $\sum \sup_{1\le s\le 2}$ principle, basically. $\endgroup$
    – FShrike
    Jun 22, 2023 at 17:53
  • 1
    $\begingroup$ Sorry I changed my mind and had deleted my comment before I saw your comment, but thanks a lot for the answer and the comment! $\endgroup$
    – hteica
    Jun 22, 2023 at 17:56

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