Proving Σ0ooarctan(1/(n2+n+1))=π/2

  • Thread starter quantumdude
  • Start date
In summary: Edited by bogdan on 12-18-2000 at 09:07 PM]In summary, the conversation was regarding a thread started by Hurkyl in PF v2.0, where they were discussing a proof for the equation Σ0ooarctan(1/(n2+n+1))=π/2. The conversation included a copy of the original thread and solutions presented by different users. One user, Dario, presented a brilliant solution using a decomposition of the fraction and a little-known formula for addition of inverse trigonometric functions. Another user, Tom, also had a solution but was hesitant to share it after seeing Dario's solution. Bogdan then presented another solution
  • #1
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,584
24
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...
 
Mathematics news on Phys.org
  • #2
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!
 
  • #3
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 ?
 
  • #4
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?
 
  • #5
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))
 
  • #6
Well, I guess I'm going to scrap my solution then.
 
  • #7
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!
 
  • #8
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:
  • #9
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:

FAQ: Proving &Sigma;0ooarctan(1/(n2+n+1))=&pi;/2

What does the formula Σ0ooarctan(1/(n2+n+1))=π/2 represent?

The formula represents the infinite sum of arctan(1/(n2+n+1)), which is equal to the value of pi divided by 2.

How is this formula derived?

This formula can be derived by using the Taylor series expansion of the arctan function and manipulating it to get the desired result.

What is the significance of this formula?

This formula has significant implications in the field of mathematics, specifically in the study of infinite series and the convergence of trigonometric series.

Can this formula be proven?

Yes, this formula has been proven by mathematicians using rigorous mathematical techniques and proofs.

Are there any real-world applications of this formula?

While this formula may not have direct real-world applications, it is a fundamental result in mathematics and has implications in other fields such as physics and engineering.

Similar threads

Back
Top