Finding P(n) for a Challenging Sum Series: A Hint for Simplifying the Expression

[FermatPell]

Homework Statement

Let

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

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

Homework Equations

$$tg(a-b)=\frac{tg(a)-tg(b)}{1+tg(a)tg(b)}$$

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 I am stuck, any hints?

 

[blather]

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 $$a$$ and $$b$$ that give us

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

Now let's find those coefficients...

 

[FermatPell]

I've tried that I the result was really ugly. Something like this:

$$a = \frac{\frac{x-1}{(k+x+1)sqrt(k+1)}+sqrt(C)}{2}$$

where C is a very complicated fraction. I don't think that's the way to go

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

MOD's please delete this post, i don't want to double post

 

[FermatPell]

Is there a simpler form of this?

$$\sum\limits_{i=2}^n \arctan(\sqrt{i})$$

 

[hunt_mat]

Care to post the solution?

 

[FermatPell]

Substitute u=sqrt(k+1), v=sqrt(k+x). After some algebra, the final solution is:

$$P(n) = (n-1) \frac{\pi}{2} - \sum\limits_{i=2}^n \arctan(\sqrt{i})$$

Now i need another way to write that

 

[blather]

That's neat! The square root of the imaginary term can be rewritten as

$$i=e^{i\tfrac{\pi}{2}}$$
$$\sqrt{i}=e^{i\tfrac{\pi}{4}}$$
$$\sqrt{i}=\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}$$
$$\sqrt{i}=\tfrac{\sqrt{2}}{2}\left(1+i\right)$$

 

[FermatPell]

No, you misunderstood it. i is not an imaginary term, its a variable (integer) that goes from 2 to n

 

[blather]

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.