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Let $$x=\lim_{n\to1^+}\left({\zeta(n)-\dfrac{1}{n-1}}\right)$$ where $\zeta$ is Riemann zeta function. What is the value of $x$?

At $n\to1^+$, $\zeta(n)\to\infty$ and $\dfrac{1}{n-1}\to\infty$, so this is indeterminate form of type $\infty-\infty$. I wrote this limit as $$x=\lim_{n\to1^+}\dfrac{\zeta(n)^2-\left({\dfrac{1}{n-1}}\right)^2}{\zeta(n)+\dfrac{1}{n-1}}$$ Then I tried to apply L'Hopital's rule, but this limit become more complicated. Also, derivative of zeta function is very complicated, so L'Hopital's rule cannot help evaluating this limit. What is the easiest way to solve it?

StubbornAtom
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  • FYI: The limit is the Euler-Mascheroni constant $\gamma \approx 0.577 \dots$ – JPhy Dec 17 '14 at 12:06
  • I know the value of limit, but I need to prove it. –  Dec 17 '14 at 12:08
  • See Equation $25$ here – Aditya Hase Dec 17 '14 at 12:08
  • For real number $x>0$ and $s\neq 1$ such that $\Re{(s)}>0$ we have $$\zeta(s)=\sum_{1 \le n \le x} \frac{1}{n^s} + \frac{x^{1-s}}{s-1}+\frac{{x}}{x^s}-s\int_x^{\infty}\frac{{u}}{u^{s+1}},du.$$ From that you can (for $x=1$) find that $$\zeta(s)=\frac{s}{s-1}-s\int_1^{\infty} \frac{{u}}{u^{s+1}},du,$$ so zeta has pole in $s=1$ and...That might help. – Cortizol Dec 17 '14 at 12:12
  • I can prove it using Hermite's formula for Hurwitz zeta function should I post my solution? – Aditya Hase Dec 18 '14 at 04:00

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