- #1
epkid08
- 264
- 1
I have two questions:
Why hasn't the hypothesis been proved yet? Is it because we don't know why re(s) has to be 1/2 and thus can't prove it, or is it because we know why re(s) has to be 1/2 but we just don't know how to prove it.
Why exactly does re(s) have to be 1/2?
[tex] \zeta (s)=1/(1-2^(1-s)) \sum_{n=0}^{\infty}1/(2^(n+1)) \sum_{k=0}^{n} (-1)^k (n ; k) (k+1)^{-s} [/tex]
If re(s) is greater than 0, and if im(s) is any real number, doesn't the function always converge to zero? In the above equation, as k approaches infinity, the denominator of the term, [tex](k+1)^{1-s}[/tex], always approaches infinity, for any value of re(s)>0, which in turn, has the series approaching zero, always. Please help me understand where I am wrong.
Why hasn't the hypothesis been proved yet? Is it because we don't know why re(s) has to be 1/2 and thus can't prove it, or is it because we know why re(s) has to be 1/2 but we just don't know how to prove it.
Why exactly does re(s) have to be 1/2?
[tex] \zeta (s)=1/(1-2^(1-s)) \sum_{n=0}^{\infty}1/(2^(n+1)) \sum_{k=0}^{n} (-1)^k (n ; k) (k+1)^{-s} [/tex]
If re(s) is greater than 0, and if im(s) is any real number, doesn't the function always converge to zero? In the above equation, as k approaches infinity, the denominator of the term, [tex](k+1)^{1-s}[/tex], always approaches infinity, for any value of re(s)>0, which in turn, has the series approaching zero, always. Please help me understand where I am wrong.
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