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Let the zeta function be defined as:

$\zeta_{A}^{B}(x) = \sum_{n=A}^{B} \frac{1}{n^{x}}$,

where $A,B \in \mathbb{Z}$ with $A\leq B$ and $x \in \mathbb{C}$.

Consider now $a_{n} = 1, 2, 4, 11, 31, 83, 227,...$ (OEIS-A002387) which is based on $\zeta_{1}^{\infty}(1)$.

Then numerical evidence indicates that:

$ e^{x-1} = \lim_{n \rightarrow \infty} ~ \frac{ \zeta_{A}^{B}(x) }{ \zeta_{B}^{C}(x) } ~~~~~~~~~~~(\text{with}~A=a_{n+0}, ~B=a_{n+1} ~\text{and}~ C=a_{n+2})$,

for all $x \in \mathbb{C}$ except those $x$ of the form $1+2ki\pi$ with $k\in\mathbb{Z}\backslash\{0\}$.

My question: Is this conjecture known or even proven/disproven ?

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    $\begingroup$ I think the result $e^{x-1}$ is correct and easy to prove, since by the definition from the link $a_{n+1}=ea_n(1+o(1))$. E.g.: put the sums in form of Riemann sums after extracting a factor, and take the limit. $\endgroup$ Jul 2, 2013 at 9:40
  • $\begingroup$ Pietro is right. Your conjecture is true and it is simple to prove along these lines. Regarding the question of whether your conjecture is known or even proven/disproven. I doubt it, since it seems too specialized. $\endgroup$ Jul 2, 2013 at 10:32
  • $\begingroup$ Note that the condition on $x$ makes the corresponding integral $\int_1^e t^{-x}dt$ different from $0$, which allows to drop it from numerator and denumerator and conclude. $\endgroup$ Jul 2, 2013 at 10:48
  • $\begingroup$ @Pietro: Thank you! So essentially, this proves the conjecture. $\endgroup$
    – user35234
    Jul 2, 2013 at 12:28

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