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I was messing around with some infinite sums in $\ell^p$ spaces and I encountered a strange result:

$\zeta\left(1 + \frac{2}{x}\right)$ looks like it is linear in $x$ for $x > 1$! A simple linear regression gives me $\zeta\left(1 + \frac{2}{x}\right)\approx 0.593413 + 0.499801 x$. The greatest difference appears to be relatively small. an attempt to determine the error

Is this true? If so, how could I show this? And if not, why does it appear to be nearly linear?

jjagmath
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NoOneIsHere
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  • Check its series expansion here. For $x \rightarrow \infty$ you have a line with slope $1/2$ and the Euler-Mascheroni constant as independent term. –  Jan 08 '22 at 00:42
  • I'm familiar with the series expansion of zeta but at least to me it's not intuitive that all but the linear order term are vanishing in that expansion (or something else I'm not aware of is going on) – NoOneIsHere Jan 08 '22 at 00:45
  • The terms with negative powers of $x$ will vanish. Just take the limit of $1/x^n$ with a positive value of $n$ and take $x \rightarrow \infty$. –  Jan 08 '22 at 00:49
  • You're welcome, NoOne! –  Jan 08 '22 at 00:53
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    We would see the same behavior for any function with a simple pole at $z=1$. – Greg Martin Jan 08 '22 at 01:48
  • $100$ is a small number. Try plotting on $[1,10000]$. – markvs Jan 08 '22 at 01:55
  • @GregMartin that's a very good point, I actually didn't know that zeta has a simple pole at $z = 1$ but that does make a lot more sense – NoOneIsHere Jan 08 '22 at 02:23

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Expanded for large values of $x$ $$\zeta \left(1+\frac{2}{x}\right)=\gamma+\frac 12 x+\sum_{n=1}^\infty (-1)^n\,\frac {2^n\,\gamma _n } {n!\,x^n }$$

For $x=1$ $$\sum_{n=1}^\infty (-1)^n\,\frac {2^n\,\gamma _n } {n! } \sim \frac 1 8$$

Edit $$\zeta (1+\epsilon)=\frac 1 \epsilon+\gamma+\sum_{n=1}^\infty (-1)^n\,\frac{\gamma _n}{n!}\,\epsilon^n$$

Claude Leibovici
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