# Thread: Series involving summation of arctan()

1. ## Series involving summation of arctan()

I am very new to this kind of thing, so I would really appreciate at least a little hint:

Main goal is to evaluate

$\displaystyle $\sum\limits_{n = 1}^{ + \infty } {\arctan \left( {\frac{1}{{2 \cdot {n^2}}}} \right)}$$

and in the process to come up with the partial summation formula.

I actually figured out the nature of adding two arctan() results

$\displaystyle $\arctan \left( x \right) + \arctan \left( y \right) = \arctan \left( {\frac{{x + y}}{{1 - x \cdot y}}} \right)$$

but still not able to move forward...

2. Put $x:=\frac 1{n-1}$ and $y:=-\frac 1{n+1}$.

3. Originally Posted by girdav
Put $x:=\frac 1{n-1}$ and $y:=-\frac 1{n+1}$.
I believe that would give me

$\displaystyle $\arctan \left( {\frac{2}{{{n^2}}}} \right)$$

so maybe a little correction would produce

$\displaystyle ${\arctan \left( {\frac{1}{{2 \cdot {n^2}}}} \right)}$$

precisely?

By the way, how do you choose this kind of substitution? I am not mentally able to comprehend it

4. Second attempt $x= \frac 1{2n-1}$ and $y = \frac 1{2n+1}$.

5. Originally Posted by girdav
Second attempt $x= \frac 1{2n-1}$ and $y = \frac 1{2n+1}$.
I believe $x= \frac 1{2n-1}$ and $$y = \frac{{ - 1}}{{2n + 1}}$$

But how to comprehend it other than come and ask in MHF forum?

I understand that that I need x and y such that

$$\frac{{x + y}}{{1 - x \cdot y}} = \frac{1}{{2 \cdot {n^2}}}$$

is true for every n. But how to know the form of x and y?

6. You can hope that you will find a telescopic series so you try to find x_n such that $\frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}$. We can try $x_n =\frac 1{an+b}$ and find $a$ and $b$.

7. Originally Posted by girdav
You can hope that you will find a telescopic series so you try to find x_n such that $\frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}$. We can try $x_n =\frac 1{an+b}$ and find $a$ and $b$.
Thanks for help. Apparently that telescopes pretty flawlessly and produces

$\displaystyle $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{\pi }{4} - \arctan \left( {\frac{1}{{2n + 1}}} \right)} \right) = \frac{\pi }{4}$$

P.S. Is anything related to

$\displaystyle $Arc\tan \left( \chi \right) = \frac{i}{2} \cdot Ln\left( {\frac{{i + \chi }}{{i - \chi }}} \right)$$

is a no-go while evaluating similar sums?