Hurkyl's Sum

  • #1
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...
 

Answers and Replies

  • #2
Hurkyl
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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!
 
  • #3
bogdan
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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
quantumdude
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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
quantumdude
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Well, I guess I'm going to scrap my solution then.
 
  • #7
Hurkyl
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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
 
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  • #9
Hurkyl
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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
bogdan
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I hate using derivatives to "compute" sums...
 
  • #12
metacristi
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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]
 
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  • #13
bogdan
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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
metacristi
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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...
 
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