Taylor expansion of f(x)=arctan(x) at infinity

I have to write taylor expansion of f(x)=arctan(x) around at x=+∞.
My first idea was to set z=1/x
and in this case z→0
Thus I can expand f(z)= arctan(1/z) near 0
so I obtain 1/z-1/3(z^3)
Then I try to reverse the substitution but this is incorrect .I discovered after that arctanx=π/2−arctan(1/x) for all x>0 and than I can write the correct result walphram alpha gives to me
Now I would like to know if there is an alternative way to get the result without knowing arctanx=π/2−arctan(1/x) for all x>0

laurabon said:

I have to write taylor expansion of f(x)=arctan(x) around at x=+∞.
My first idea was to set z=1/x
and in this case z→0
Thus I can expand f(z)= arctan(1/z) near 0
so I obtain 1/z-1/3(z^3)

How did you obtain this result? Taking the series for [itex]\arctan t[/itex] about [itex]t = 0[/itex] and setting [itex]t = z^{-1}[/itex] won't give the right answer.

This idea will work, if you calculate the series correctly.

For the constant term, [tex]f(0^{+}) = \arctan(\infty) = \frac{\pi}2.[/tex] Then [tex]f'(z) =
\frac{d}{dz}\arctan(z^{-1}) = -\frac{1}{z^2} \frac{1}{1 + z^{-2}} = - \frac{1}{1 + z^2}[/tex] and then [tex]
f^{(n)}(z) = \frac{d^n}{dz^n} \arctan(z^{-1}) = -\frac{d^{n-1}}{dz^{n-1}} \frac{1}{1 + z^2},\qquad n \geq 2.[/tex] Alternatively, you could note that if [itex]x = \tan \theta[/itex] then [itex]x^{-1} = \cot \theta[/itex] so that [itex]f(z) = \operatorname{arccot}(z)[/itex].

Hi , thanks for your replies . Can I use the fact that 1/(1-x)=1+x+x^2+...
and so 1/(1+x^2)=1-x^2+x^4+...
and obtain f'(0)=-1 f^3(0)=-2! and so on ?