## Riemann Zeta Function

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tomtom2357
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### Riemann Zeta Function

I was playing around with summation, and I found this:

Sumn=2infinity(1/(n(n-1))
=Sumn=2infinity(1/(n^2)+1/(n^3)+1/(n^4)+...)
=Sumn=2infinity(Sumi=2infinity(1/(n^i)))
=Sumi=2infinity(Sumn=2infinity(1/(n^i)))
=Sumi=2infinity(Z(i)-1), where Z is the Riemann Zeta Function.

But Sumn=2x(1/(n(n-1))=(x-1)/x
So Sumn=2infinity(1/(n(n-1))=1
Therefore Sumi=2infinity(Z(i)-1)=1

Is this right?
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Heptadecagon
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### Re: Riemann Zeta Function

It does seem right, all the steps are good, and I looked it up online , and it is a known fact. But a very cool thing to discover! Thanks for sharing!

tomtom2357
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### Re: Riemann Zeta Function

Where did you find out that it was a known fact? I have a few other ideas similar to that one, and I want to check them too. Thanks for confirming my reasoning!
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mr-mitch
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### Re: Riemann Zeta Function

tomtom2357
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### Re: Riemann Zeta Function

I have another identity, that is not on the list:
Sumi=3infinity(i-1)(Z(i)-1)=1
Proof on request
Spoiler:
Expand 1/(x^2-2x+1)

Could this be used to derive bounds on the riemann zeta function?
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

Sagekilla
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### Re: Riemann Zeta Function

Another fun identity is that the sum in the OP equates to 1.
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tomtom2357
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### Re: Riemann Zeta Function

One think I realized though, is that you have to be extremely careful when calculating sums like this, or you end up getting contradictions when the sums don't converge.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

tomtom2357
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### Re: Riemann Zeta Function

I discovered another interesting identity: Sum(i=n+1 to infinity: 1/((i2)-ni))=Hn/n). It seems to work for small n, but I'm not sure how to prove it.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

jestingrabbit
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### Re: Riemann Zeta Function

tomtom2357 wrote:I discovered another interesting identity: Sum(i=n+1 to infinity: 1/((i2)-ni))=Hn/n). It seems to work for small n, but I'm not sure how to prove it.

Do partial fractions on the summand.
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tomtom2357
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### Re: Riemann Zeta Function

I came up with another interesting identity (and I didn't know where to put it, so I put it here): Sum n=0 to infinity (rncos(an))=(1-r*cos(a))/(1-2r*cos(a)+r2), and Sum n=0 to infinity (rnsin(an))=r*sin(a)/(1-2r*cos(a)+r2) (at least when |r|<1). You can prove it using eix=cos(x)+i*sin(x), and the geometric series. Has this been done before?
Last edited by tomtom2357 on Tue Oct 22, 2013 1:56 am UTC, edited 2 times in total.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

dudiobugtron
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### Re: Riemann Zeta Function

tomtom2357 wrote:I came up with another interesting identity (and I didn't know where to put it, so I put it here): Sum n=0 to infinity (rncos(an))=(1-r*cos(a))/(1-2r*cos(a)+r2), and Sum n=0 to infinity (rnsin(an))=r*sin(a)/(1-2r*cos(a)+r2) (at least when |a|<1). You can prove it using eix=cos(x)+i*sin(x), and the geometric series. Has this been done before?

I don't know enough to address or even properly understand your question before I need to go back to cooking dinner; I just thought I'd check - did you mean to write "sum i=0...", or "sum n=0..."? It seems like a pretty amazing identity if you did actually mean i.

tomtom2357
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### Re: Riemann Zeta Function

dudiobugtron wrote:
tomtom2357 wrote:I came up with another interesting identity (and I didn't know where to put it, so I put it here): Sum n=0 to infinity (rncos(an))=(1-r*cos(a))/(1-2r*cos(a)+r2), and Sum n=0 to infinity (rnsin(an))=r*sin(a)/(1-2r*cos(a)+r2) (at least when |a|<1). You can prove it using eix=cos(x)+i*sin(x), and the geometric series. Has this been done before?

I don't know enough to address or even properly understand your question before I need to go back to cooking dinner; I just thought I'd check - did you mean to write "sum i=0...", or "sum n=0..."? It seems like a pretty amazing identity if you did actually mean i.

Huh? When I said sum, I meant the summation of rncos(an) over all the non-negative integers n (for the first sum). The expanded first sum is: r0cos(0)+r1cos(a)+r2cos(2a)+...
The expanded second sum is:
r0sin(0)+r1sin(a)+r2sin(2a)+...
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

dudiobugtron
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### Re: Riemann Zeta Function

OK cool. I probably should have also said that I edited your 'i's to 'n's. I bolded them to show it, but it's not at all obvious. tomtom2357
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### Re: Riemann Zeta Function

Oh, right. I'll edit my post. It was supposed to be sum n=0, thanks for pointing that out.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

cyanyoshi
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### Re: Riemann Zeta Function

Those formulas are indeed quite nifty. I verified them and they appear to be correct! Sorry to nitpick, but I think you meant that the sums are defined (in the usual sense) only when |r|<1, not |a|<1. You aren't the first to figure this out. The formulas are given in Exponential Sum Formulas. (Look at the last sentence.) I encountered some sums of this form just a week or two ago when working through some structural dynamics problems, but we were fortunately allowed to just calculate them numerically.

tomtom2357
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### Re: Riemann Zeta Function

I think they're quite cool because I've come across soecific examples of this sum before, and I always wondered what the sum was equal to, but all that was said was that it converged (if |r|<1).
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.

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