Question

please Solve this it is important
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Answer 1

Answer:

Let ζ(s)=∑∞n=11ns a standard formula.

I'm confused if you tell me: does this series: ∑∞n=11ns converge?

I will answer you: this series is divergent. But if you say: ζ(−2) it will be: ζ(−2)=∑∞n=11n−2=0. Will be convergent. So why ?

Similar questions has been asked before, see for example this. Note that this link talks about ζ(−1), but for what is the essence of your questions this is the same thing.

The fact that the analytic continuation has a finite value (namely, 0) does not vouch for convergence of the series. A simpler example: 1+2+22+23+… arguably blows up, yet this is the analytic continuation to x=2 of 1/(1−x), so the analytic continuation takes value −1. Hard to rationalize how the sum of positive reals could be negative. (Hilariously, in the 2-adics, that conclusion is literally correct, but that is tangential to the present issue.)

When people say "zeta function has zeros on negative even integers", they are talking about the analytic continuation of the Riemann zeta function, and the naive formula only works for Re(s)>1

See this, which gives a functional equation for ζ(s):

ζ(s)=2sπs−1sin(πs2)Γ(1−s)ζ(1−s)

All negative even integers are "trivial zeros" of the zeta function, because sin(πs2) would equal 0.

Step-by-step explanation:

Hope this helps you

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