Proving Σ0ooarctan(1/(n2+n+1))=π/2

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Discussion Overview

The discussion revolves around the proof of the series sum Σ0ooarctan(1/(n^2+n+1))=π/2. Participants explore various approaches to derive this result, including parametrization, trigonometric identities, and integration techniques. The conversation includes both theoretical insights and mathematical reasoning.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes to parametrize the sum and differentiate to simplify the arctan function.
  • Another participant suggests a solution using the tangent subtraction formula, leading to a conclusion that the sum equals π/2.
  • A participant questions the validity of a step in the tangent subtraction derivation.
  • Some participants express uncertainty about their own solutions and the use of derivatives in summation.
  • Another approach involves rewriting arctan using discrete math and trigonometric identities, leading to a limit that also suggests the sum converges to π/2.
  • One participant acknowledges the complexity of spotting the solution and presents it as a challenge problem.
  • There is a discussion about the continuity and behavior of the function involved in the integral approach to the sum.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the validity of various approaches. While some solutions are praised, others are met with skepticism or requests for clarification. No consensus is reached on a single definitive method for proving the sum.

Contextual Notes

Some participants note limitations in their approaches, such as the need for clearer definitions or the potential for errors in reasoning. The discussion reflects ongoing exploration rather than settled conclusions.

Who May Find This Useful

Readers interested in advanced mathematical proofs, series convergence, and trigonometric identities may find the discussion valuable.

quantumdude
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Congratulations Hurkyl, you're famous.

This was a thread that he started in PF v2.0, and which I participated in. I made a copy of it because I left it unfinished.

Prove:

Σ0ooarctan(1/(n2+n+1))=π/2

My (as yet incomplete) solution:

Parametrize the sum as follows:
S(a)=Σ0ooarctan(a/(n2+n+1))
S'(a)=Σ0oo(n2+n+1)/(a2+(n2+n+1)2)

That gets rid of the nasty arctan function.

My approach will be to find the sum of S'(a) and integrate with respect to 'a' with the limits 0<a<1. That will give me:
S(1)-S(0)=S(1), which is the original sum.

To be continued...
 
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I had forgotten I put that here!

That new sum still doesn't look pretty, but that doesn't mean it can't be summed! I'll be watching to see where you go from there!
 
I think I have a solution...

1/(n^2+n+1)=(n+1-n)/[1+n(n+1)]...

Let tan(a) = n+1; tan(b) = n;

arctan(1/(n^2+n+1))=arctan(tan(a)-tan(b)/(1+tan(a)*tan(b)))=
=arctan(tan(a-b))=a-b=arctan(n+1)-arctan(n)...

So arctan(1/(n^2+n+1))=arctan(n+1)-arctan(n)...
Suming we obtain sum = arctan(infinity)-arctan(0) = pi/2...

Is it good ?
 
Originally posted by bogdan
arctan(tan(a)-tan(b)/(1+tan(a)*tan(b)))=
=arctan(tan(a-b))

How did you jump from the first line to the second?
 
Originally posted by bogdan
I think I have a solution...
...
Is it good ?
[/B]

IT SURE IS! AWESOME, BRILLIANT solution, actually!

My compliments, Dario

Originally posted by Tom
How did you jump from the first line to the second?

The formula used is simply the one for the tangent of a sum(difference) of arcs
tan(a-b)=(tan(a)-tan(b))/(1+tan(a)*tan(b))
 
Well, I guess I'm going to scrap my solution then.
 
Bah don't give up!

I've seen at least one pretty solution that didn't use that cheap trick, it would be interesting to see another one!
 
Originally posted by Hurkyl
Bah don't give up!

I've seen at least one pretty solution that didn't use that cheap trick, it would be interesting to see another one!

The solution is actually good and brilliant since it uses a good hint of insight for the decomposition of the fraction and a pretty much unknown formula for addition of inverse trigonometric functions but if you want to give us your definition of "cheap trick" I am curious to read it!

Said that, I am sure that Tom's approach can take somewhere at least the idea to focus on a derivate sum and integrate. I am not sure about the functional form he has chosen but it is a possibility...

Dario
 
Last edited by a moderator:
I was just teasing. :smile: I guess I didn't make it all that clear, though.

I derived the sum via that very argument, and I present it as a challenge problem because I know how difficult it would be to spot.
 
  • #10
gotcha ;)
 
  • #11
I hate using derivatives to "compute" sums...
 
  • #12
Bogdan's great solution remembered me that arctg[1/(n2+n+1)] can be convenably rewritten using some discrete math considerations and trigonometry also.

If dx is a positive variation (not necessarilly infinitesimal) around a parameter x then we have:

arctan(x+dx)-arctan(x)=arctan[dx/(1+x*dx+x2)] (1)

where I used the trigonometric formula

arctan(a)-arctan(b)=arctan[(a-b)/(1+a*b)] (2)

If we replace in (1) x with n and put the condition that the step is dx=1 --->

arctan(n+1)-arctan(n)=arctan[1/(1+n+n2)] (3)

After simplifying the terms the initial sum can be rewritten as:

S=limk->00∑;from n=0 to k arctan[1/(1+n+n2)]

S=limk->00arctan[k+1]=arctan(00)=π/2.

Of course I've written all these only in order to check some HTML mathematical symbols :-).

[edit to correct some gramatical mistakes]
 
Last edited:
  • #13
Yeap...
arctan(dx/(1+x*dx+x^2))/dx=1/(1+x^2)...
because...
arctan(dx/(1+x*dx+x^2))/dx=arctan(dx/(1+x*dx+x^2))/(dx/(1+x*dx+x^2))*1/(1+x*dx+x^2)=1*1/(1+x^2)...because lim arctanx/x=1, x->0...
So...we obtain the "derivative" for arctan...
 
  • #14
Tom

I've just realized that you can also find the sum S of the series &#8721n=0&#8734 arctan[1/(n2+n+1)] by calculating the integral:

I=&#8747x=0&#8734 arctan[1/(x2+x+1)]

Indeed the function f(x)=arctan[1/(x2+x+1)] is continuous and decreasing over the interval [0,&#8734] and f(1)=term[1],f(2)=term[2] and so on.I is the seeked sum when n->&#8734.

[edit to add]

Of course I am wrong.At most the comparation with the above integral proves that the series is convergent...
 
Last edited:

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