Re: sum of zetas
- From: "pi" <Jason628@xxxxxxxxx>
- Date: 7 Sep 2005 03:16:26 -0700
Gottfried Helms wrote:
> Can the sum
>
> x = sum zeta(2^(2^i) ) = zeta(2) + zeta(4) +zeta(16) + zeta(256) + ...
> i=0
>
> be given in a more simple form in common functions?
>
> Gottfried Helms
The Zeta functions approaches 1 hence, your sum becomes
zeta(2) + zeta(4) +zeta(16) + zeta(256) + ... + zeta(2^(2^m))
(m+1).727272582818773279860151434674829814016938242115996876571639648402915173618131508345379235852406471
for instance if m= 4 then the sum is
5.727272582818773279860151434674829814016938242115996876571639648402915173618131508345379235852406471
if m=5 the sum is
6.727272582818773279860151434674829814016938242115996876571639648402915173618131508345379235852406471
.
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