# A limit involving Riemann zeta function $\zeta(s)$

I am reading Apostol's "Introduction to Analytic Number Theory". In Chapter 3, he defines Riemann Zeta Function, $$\zeta(s)$$, by the equation $$$$\zeta(s) = \sum_{n = 1}^{\infty} \frac{1}{n^s} \quad \text{if } s > 1,$$$$ and by the equation $$$$\zeta(s) = \lim_{x \to \infty} \left( \sum_{n \leq x} \frac{1}{n^s} - \frac{x^{1 - s}}{1 -s} \right) \quad \text{if } 0 < s < 1.$$$$ Then he derives the following asymptotic formulas. $$$$\label{harmonic_partial_sums} \sum_{n \leq x} \frac{1}{n} = \log x + C + O\left(\frac{1}{x}\right),$$$$ $$$$\label{1/ns partial sums} \sum_{n \leq x} \frac{1}{n^s} = \zeta(s) + \frac{x^{1 - s}}{1 - s} + O(x^{-s}),$$$$ where $$C$$ is the Euler-Mascheroni constant. Looking at these formulas, one might suspect (naively) that $$$$\lim_{s \to 1} \left( \zeta(s) + \frac{x^{1 - s}}{1 -s} \right) = \log x + C.$$$$ I tried to prove this by going back to the Abel summation formula but came across the integral $$$$\int_{x}^{\infty} \frac{\{t\}}{t^{s+1}} \, dt$$$$ whose limit does not exist as $$s$$ approaches $$1$$. I would highly appreciate if I could know whether the limit exists and if it exists what's its value.

• The limit actually exists when $s\to1$ from the right. Commented Jul 19, 2022 at 7:44

The Riemann zeta function has a simple pole at $$s=1$$, with residue $$1$$. If you take $$x\gt0$$ a real number, as I think you are, then: \begin{align}\lim_{s\to1}(\zeta(s)+x^{1-s}(1-s)^{-1})&=\lim_{s\to1}\left[\frac{x^{1-s}-1}{1-s}+\underset{\longrightarrow\gamma}{\underbrace{(\zeta(s)-(s-1)^{-1})}}\right]\\&\overset{L.H}=\gamma+\lim_{s\to1}\frac{(-\ln x)x^{1-s}}{-1}\\&=\ln(x)+\gamma\end{align}

As you expected, where the constant $$C$$ is the $$\gamma$$ the Euler-Mascheroni constant, or the first Stieltjes constant.

To see where these constants are coming from, I will reference for you an answer someone gave me a few months ago, here. No complex analysis is involved, really - you should only understand that "analytic continuation is unique" to understand why Conrad's zeta is the same as your zeta.

$$\gamma=\lim_{n\to\infty}\left[-\ln n+\sum_{m=1}^n\frac{1}{m}\right]$$

• Is there any elementary method to show the limit $\lim_{s \to 1} (\zeta(s) - (s-1)^{-1}) = \gamma$.
– Ryan
Commented Jul 19, 2022 at 8:49
• @MuhammadAtifZaheer Please see the answer I linked, it is as elementary as it could reasonably get. The thing is about the zeta function, there are a gazillion different representations, all equivalent on various different domains... sometimes other representations, like Conrad's, are more useful than Apostol's Commented Jul 19, 2022 at 8:49
• @MuhammadAtifZaheer The main problem with what? Commented Jul 19, 2022 at 9:51
• I think the main problem is with interchanging the following limits: $\lim_{s \to 1} \lim_{x \to \infty} \sum_{n \leq x} \frac{1}{n^s} + \frac{x^{1-s} - 1}{s - 1} = \lim_{x \to \infty} \lim_{s \to 1} \sum_{n \leq x} \frac{1}{n^s} + \frac{x^{1-s} - 1}{s - 1}$.
– Ryan
Commented Jul 19, 2022 at 9:53
• The left hand side is $\lim_{s \to 1} \left \{ \zeta(s) - \frac{1}{s - 1} \right \}$ and the right hand side is $\gamma$.
– Ryan
Commented Jul 19, 2022 at 9:57