- #1
kezman
- 37
- 0
Can somebody explain me about the trivial zeros?
Why [tex] \zeta(-2) = \zeta(-4) = \zeta(-6) = 0 = \zeta(k)[/tex]
So [tex] \zeta(k) \sum_{n=1}^{ \infty} \frac{1}{n^k} = 0 [/tex]?
Why [tex] \zeta(-2) = \zeta(-4) = \zeta(-6) = 0 = \zeta(k)[/tex]
So [tex] \zeta(k) \sum_{n=1}^{ \infty} \frac{1}{n^k} = 0 [/tex]?