Timeline for Prove that the real part of this limit converges to $\frac{1}{2}$
Current License: CC BY-SA 4.0
4 events
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| Aug 20, 2023 at 20:39 | comment | added | Mats Granvik |
(*start*)(*Mathematica 8.0.1*) n = 40;(*set n=40 for more digits*) s = (1 + 14*I); x = 1/10^20; N[-s - 1/n - 1/(1 - Sum[(-1)^(k - 1)* Binomial[n - 1, k - 1]/((s + k/n) Zeta[1 + s + k/n] x), {k, 1, n}]/Sum[(-1)^(k - 1)* Binomial[n - 1, k - 1]/((s + k/n + 1/n) Zeta[1 + s + k/n + 1/n] x), {k, 1, n}]), n/2](*end*)
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| Mar 19, 2023 at 8:40 | comment | added | Mats Granvik | This is analytic continuation after all, although I did not think so at first. | |
| Feb 13, 2023 at 14:42 | comment | added | Mats Granvik |
$s= 3/2 + 14i \\ M=10^{10000}$ $$\Re\lim_{n \rightarrow 100} \left( \left[ 1- \left( \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s\right)}} \Bigg/ \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\sum_{m=1}^{m=M} 1/m^{\left(\tfrac{k}{n}+s+\tfrac{1}{n}\right)}} \right) \right]^{-1} +\frac1n + s \right) = $$ 0.49999999999999999999999999999999999999999999999999999999999999999999\ 821263876336076842 + 14.134725141734693790457251983562470270784257115699243175685567460149\ 959971757841139084 I
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| Feb 13, 2023 at 14:08 | history | answered | Mats Granvik | CC BY-SA 4.0 |