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Old 2019-03-20, 16:34   #1
mart_r
 
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Default sum (1/n^s), negative values of

On the wing of an interest in the zeta function, I've been working on this conundrum in the past two days:


Find the largest \delta such that there exists x and \tau such that
\sum_{n=1}^x \frac{1}{n^{1+\delta+i \tau}}<0


The idea behind it is while \zeta(\sigma+i\tau) is >0 for \sigma>1, the intermediate summation terms can still be <0. What is the largest \sigma for this to happen?


(My state of knowledge is that [TEX]\delta[/TEX] > 1/8 if I did the math right.)

Last fiddled with by mart_r on 2019-03-20 at 16:42
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Old 2019-03-21, 19:04   #2
Dr Sardonicus
 
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Quote:
Originally Posted by mart_r View Post
On the wing of an interest in the zeta function, I've been working on this conundrum in the past two days:


Find the largest \delta such that there exists x and \tau such that
\sum_{n=1}^x \frac{1}{n^{1+\delta+i \tau}}<0


The idea behind it is while \zeta(\sigma+i\tau) is >0 for \sigma>1, the intermediate summation terms can still be <0. What is the largest \sigma for this to happen?


(My state of knowledge is that [TEX]\delta[/TEX] > 1/8 if I did the math right.)
The sums you indicate are likely to be non-real. Perhaps you mean the real part?

The following might be pertiment: A NOTE ON THE REAL PART OF THE RIEMANN ZETA-FUNCTION
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Old 2019-03-21, 22:03   #3
mart_r
 
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Quote:
Originally Posted by Dr Sardonicus View Post
The sums you indicate are likely to be non-real. Perhaps you mean the real part?

The following might be pertiment: A NOTE ON THE REAL PART OF THE RIEMANN ZETA-FUNCTION
That looks just like the thing I was looking for.
So \zeta(\sigma+i\tau) can be <0 when \sigma>1 - must have gotten out of my sight.

Thanks a lot!
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