A challenging sum series

  • Thread starter FermatPell
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  • #1
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Homework Statement


Let

[tex]P(x) = \displaystyle\sum\limits_{k=1}^\infty arctan (\frac{x-1}{(k+x+1)sqrt(k+1) + (k+2)sqrt(k+x)})[/tex] (infite sum from K=1 to infinity)

a) Simplify the expression for P(n), where n is a non-negative integer

Homework Equations



[tex]tg(a-b)=\frac{tg(a)-tg(b)}{1+tg(a)tg(b)}[/tex]

The Attempt at a Solution



Im looking for a hint. My first idea was to somehow use trig formulae, but i can't factorize the denominator or split it into partial fractions. Now im stuck, any hints?
 

Answers and Replies

  • #2
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I think I sort of see how to do this...Let's give it a go. I'll post more later if something else comes to me:

First, there's that pesky arctangent in there. So, let's use the relation you have to get rid of it. That is, we need to propose two variables [tex]a[/tex] and [tex]b[/tex] that give us

[tex]\Sigma\arctan[\tan(a-b)]=\Sigma(a-b)[/tex]

Now let's find those coefficients...
 
  • #3
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I've tried that I the result was really ugly. Something like this:

[tex]a = \frac{\frac{x-1}{(k+x+1)sqrt(k+1)}+sqrt(C)}{2}[/tex]

where C is a very complicated fraction. I dont think thats the way to go

EDIT: I've succeeded. :D :D but followed a different path..

MOD's please delete this post, i dont want to double post
 
Last edited:
  • #4
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Is there a simpler form of this?

[tex]$\sum\limits_{i=2}^n \arctan(\sqrt{i})$[/tex]
 
  • #5
hunt_mat
Homework Helper
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Care to post the solution?
 
  • #6
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Substitute u=sqrt(k+1), v=sqrt(k+x). After some algebra, the final solution is:

[tex]P(n) = (n-1) \frac{\pi}{2} - \sum\limits_{i=2}^n \arctan(\sqrt{i})[/tex]

Now i need another way to write that
 
  • #7
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That's neat! The square root of the imaginary term can be rewritten as

[tex]i=e^{i\tfrac{\pi}{2}}[/tex]
[tex]\sqrt{i}=e^{i\tfrac{\pi}{4}}[/tex]
[tex]\sqrt{i}=\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}[/tex]
[tex]\sqrt{i}=\tfrac{\sqrt{2}}{2}\left(1+i\right)[/tex]
 
  • #8
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No, you misunderstood it. i is not an imaginary term, its a variable (integer) that goes from 2 to n
 
  • #9
33
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Ah.

Seems like you've got this one. I'm curious, could you post the solution when you get it? I'd be much obliged.
 
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