Power series representation for (x^2+1)*arctan(x)

[Amit_Davidi]

Hello everybody!
i've been stuck on this problem for some time now, and i can't figure out how to place (X^2 + 1) inside the summation of arctan(x)

(1+x^2) * Sigma : [ (-1^)n * x^2n+1 ] / 2n+1
any help would be greatly appreciated my friends !

 

[tkhunny]

"place inside" is a troubling expression. Are you SURE it means anything?

Multiplying infinite series generally falls into an odd category that some like to call "Essentially Impossible". In other words, you can sort of do it and it may lead to something, but it is typically seen as far too much trouble for any value that might be garnered. In some cases, results of some value can be obtained.

This one is not so bad, since one of the series is finite:

[math]1\cdot\left(x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7} + \dfrac{x^{9}}{9} - \;etc\right) +\;x^{2}\cdot\left(x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7} + \dfrac{x^{9}}{9} - etc\right)[/math]
Leads to:

[math]x(1) + x^{3}(1-1/3) - x^{5}(1/3 - 1/5) + x^{7}(1/5 - 1/7) - x^{9}(1/7 - 1/9) - etc[/math]or
[math]x + x^{3}(2/3) - x^{5}(2/15) + x^{7}(2/35) - x^{9}(2/63) + etc[/math]
Your only remaining challenge is that denominator. It looks like ((2n)^2 - 1), excepting that first term.

Let's see what you get. Then, let's see you worry about convergence.

 

[Amit_Davidi]

So i got [MATH]\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{2n+1}[/MATH]and the Convergence Radius should be the minimal of the two series.
i thought i should have one Sum expression so i got stuck on that, thank you !

 

[Amit_Davidi]

i can't quite figure out how to procceed,
[MATH]x+\:\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n\:\cdot 2x^{2n+1}}{\left(2n\right)^2-1}[/MATH]