f(x)=x/(9 + x^2)
We know that 1/(1-x) = (sigma)(0 to infinity)x^n
find f(x) and the interval of convergence.
Hello,
The most direct way is to say that :
$\displaystyle \frac{x}{9+x^2}=\sum_{n \geq 0} a_n x^n$
And to solve for $\displaystyle a_n$
Now your result... It's ok.
What mathstud meant is to observe... You know the power series for $\displaystyle \arctan x$, whose derivative is $\displaystyle \frac{1}{1+x^2}$
From here, you can note that your function is x multiplied to what looks like the derivative of an arctan.
Once you've got your power series, derivate it