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I used some graphs but did not find the correct value,

$$\sum_{n=1}^ \infty \frac{1}{n^x} = 2 $$

what is the x value?

Talha ŞAHİN
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  • $x\approx 1.7286472389981836181351030102976914642$ – Hagen von Eitzen Oct 25 '18 at 17:53
  • $x\approx 1+{\rm arccot} \left({\frac {\zeta \left( 3 \right) }{55}}+\ln \left( 3 \right) \right) $ – Mariusz Iwaniuk Oct 25 '18 at 17:57
  • @MariuszIwaniuk , thanks for approximation, but what I wonder is this: if we consider 2 as y value, then do we have a general formula? for example if it was 3, what would be changed? – Talha ŞAHİN Oct 25 '18 at 18:00
  • No it's not a general formula. Another approximation: $x\approx 1+\arctan \left( {\frac {276,\pi}{137}}-2,{\rm e} \right)$ – Mariusz Iwaniuk Oct 25 '18 at 18:05
  • @MariuszIwaniuk by the way your first approximation I guess has fault, right? Or you said radian or grad instead degree? – Talha ŞAHİN Oct 25 '18 at 18:11
  • Note $\zeta(x) = 2,x> 1$ is important because $x$ is the abscissa of convergence of $\sum_{n=1}^\infty n^{-s} c_n$ where $c_n$ is the number of factorization of $n$ in products of integers $\ge 2$ where the order counts. – reuns Oct 26 '18 at 17:06

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In order to solve $\zeta(x)=2$ we may use $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(s)=\frac{1}{s-1}+\gamma+o(1)$ for $s\to 1^+$.
$\zeta(s)$ is decreasing and convex on $(1,2)$, hence Newton's method $$ x \mapsto x-\frac{\zeta(x)-2}{\zeta'(x)} $$ with starting point $x=2$ (or, better, $\frac{3-\gamma}{2-\gamma}$) converges pretty fast to $1.72865\ldots$

Jack D'Aurizio
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