# Arc tan sum math help

1. Jul 3, 2008

1. The problem statement, all variables and given/known data
Find the sum
Arc(tan1/2)+Arc(Tan1/8)+...+Arc(Tan1/2*n^2)

2. Relevant equations

nothing

3. The attempt at a solution

2. Jul 3, 2008

Re: sum

People still haven't gotten this part yet?

Hadi, what have you attempted so far? As always, we don't do homework for you; you must show some effort before we do

Is this an infinite sum?

3. Jul 3, 2008

### HallsofIvy

Staff Emeritus
Re: sum

And don't you mean Arctan(1/2), etc. rather than Arc(tan(1/2))- else you will need to define "Arc" for me!

4. Jul 4, 2008

Re: sum

How can I edit this one

5. Jul 4, 2008

### dirk_mec1

Re: sum

You can't. Just post again this time with the correct sum.

6. Jul 4, 2008

### morphism

Re: sum

Presumably the sum is
$$\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{1}{8}\right) + \cdots + \arctan\left(\frac{1}{2n^2}\right).$$

If so, then what exactly does it mean to "find" this sum? If the goal is to simplify it, then this problem is similar to an old, well-known one that asks for a simplification of the sum
$$\sum_{k=1}^{n} \arctan\left(\frac{1}{1+k+k^2}\right).$$

One of the ways of doing this is to first note that $\arctan(k+1) - \arctan(k) = \arctan(1/(1+k+k^2))$*, and then telescope.

If you can figure out how to get this identity, then you can play around to come up with a similar one that will solve your problem.

It's also interesting to try to evaluate
$$\sum_{k=1}^{\infty} \arctan\left(\frac{1}{2k^2}\right).$$

(* What's up with [itex]?)

7. Jul 5, 2008

Re: sum

thanks a lot my question is solved