Series involving summation of arctan()

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

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...
April 13th 2011, 10:12 AM
girdav

Put $x:=\frac 1{n-1}$ and $y:=-\frac 1{n+1}$ .
April 13th 2011, 10:23 AM
Pranas

Quote:

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 :(
April 13th 2011, 10:28 AM
girdav

Second attempt $x= \frac 1{2n-1}$ and $y = \frac 1{2n+1}$ .
April 13th 2011, 10:36 AM
Pranas

Quote:

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? :D

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?
April 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 $\frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}$ . We can try $x_n =\frac 1{an+b}$ and find $a$ and $b$ .
April 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 $\frac{x_n-x_{n+1}}{1+x_nx_{n+1}}=\frac 1{2n^2}$ . 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?