# Series involving summation of arctan()

• Apr 13th 2011, 10:07 AM
Pranas
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 \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 \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...
• Apr 13th 2011, 10:12 AM
girdav
Put $\displaystyle x:=\frac 1{n-1}$ and $\displaystyle y:=-\frac 1{n+1}$.
• Apr 13th 2011, 10:23 AM
Pranas
Quote:

Originally Posted by girdav
Put $\displaystyle x:=\frac 1{n-1}$ and $\displaystyle y:=-\frac 1{n+1}$.

I believe that would give me

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

so maybe a little correction would produce

$\displaystyle \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 :(
• Apr 13th 2011, 10:28 AM
girdav
Second attempt $\displaystyle x= \frac 1{2n-1}$ and $\displaystyle y = \frac 1{2n+1}$.
• Apr 13th 2011, 10:36 AM
Pranas
Quote:

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

I believe $\displaystyle x= \frac 1{2n-1}$ and $\displaystyle $y = \frac{{ - 1}}{{2n + 1}}$$

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

I understand that that I need x and y such that

$\displaystyle $\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?
• Apr 13th 2011, 10:51 AM
girdav
You can hope that you will find a telescopic series so you try to find x_n such that $\displaystyle \frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}$. We can try $\displaystyle x_n =\frac 1{an+b}$ and find $\displaystyle a$ and $\displaystyle b$.
• Apr 13th 2011, 12:48 PM
Pranas
Quote:

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

Thanks for help. Apparently that telescopes pretty flawlessly and produces

$\displaystyle \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 \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?