How does one perform an infinite series expansion of a function "at" infinity? Question speaks for itself.
Ive seen it suggested that we substitute $x$ for $\dfrac{1}{x}$, and expand $f(\dfrac1x)$ about $0$. But the problem is, this isnt always possible, I dont think.
There is an infinite series expansion of $\arctan(x)$ at infinity that Wolfram|Alpha offers up, but when you ask it to expand $\arctan(1/x)$ at $0$, it bugs up and doesnt produce.
So Im confused about the actual approach to determining such a series.
Similar questions have been asked on SE but none ask explicitly on how.  Answers given to those questions also dont elaborate on any method.  Ive also seen reference to Laurent series and Asymptotic behavior.
 A: *

*Iff $f(1/z)$ is complex analytic on $0<|z|<r$ there is a Laurent series expansion of $f$ at $\infty$. Most other kind of expansions are obtained the same way, from $f(1/g(z))$ or from a sequence of analytic functions approximating $f$. 

*For $|z| < 1$ : $2i\text{ arctan}'(z) = \frac{2i}{1+z^2} = \frac{1}{z+i}-\frac{1}{z-i}$ so $2i\text{ arctan}(z)  = \log(z+i)-\log(z-i)$. But there are many possible branches for $\log(z+i)$ and $\log(z-i)$.Choosing the branch cut to be $[-i, i]$ it is analytic for $|z| > 1$ and hence has a Laurent series at $\infty$ valid for $|z| > 1$. But obviously mathematica implements $\text{ arctan}(z)$ as being analytic on $|z| < 1$, in this setting the branch cut is $[i,i\infty) \cup [-i,-i\infty)$ so that $\text{ arctan}(z)$ is not analytic on $|z| > 1$ and hence has no Laurent series at $\infty$.

*If $g$ is complex analytic on $r-a< |z| < r+b$ and if you like Fourier series, then compute the Fourier coefficients to obtain $$g(r e^{2i \pi n t}) = \sum_{n=-\infty}^\infty c_n e^{2i \pi n t}=\sum_{n=-\infty}^\infty \frac{c_n}{r^n} r^n e^{2i \pi n t}$$
since $g$ is complex analytic, this series is valid for $t$ complex, ie. you'll obtain that for $r-a< |z| < r+b$
$$g(z) = \sum_{n=-\infty}^\infty \frac{c_n}{r^n} z^n$$
A: Part of the problem is that Wolfram|Alpha is free to use and hence also limited in its functionality. If you were to use Mathematica, it would have had no problems with a series expansion of the form "Series[ArcTan[1/x],{x,0,10}]" and it would give you the same result as the expansion at infinity (Series[ArcTan[x],{x,0,10}]). There would only be a slight difference, which has to do with the fact that Mathematica treats the variable $x$ as a complex number, but that could be solved without to much difficulty.
So how to do the expansion of a function $f(1/x)$ in the limit $x \rightarrow 0$? It is not as difficult as it seems. You can make a Taylor expansion of the function at $x=a>0$ in some small parameter $\epsilon$
$$
f\left(\frac{1}{a+\epsilon}\right) \approx f\left(\frac{1}{a}\right) + f'\left(\frac{1}{a}\right) \epsilon + \frac{1}{2} f''\left(\frac{1}{a}\right) \epsilon^2 + {\cal O}(\epsilon^3)
$$
For this we only need to compute the various derivatives of the function we have and it would be a correct Taylor expansion at any location $x=a>0$. If we now take the limit $a \downarrow 0$ we get an expansion:
$$
f\left(\frac{1}{\epsilon}\right) \approx  \lim_{a \downarrow 0} \left\{f\left(\frac{1}{a}\right) + f'\left(\frac{1}{a}\right) \epsilon + \frac{1}{2} f''\left(\frac{1}{a}\right) \epsilon^2 + {\cal O}(\epsilon^3) \right\}  \epsilon^2 + \dots
$$
Since we want to know the limiting behaviour of $f(x)$ for large $x$ we make the identification $x=\frac{1}{\epsilon}$ and $b=\frac{1}{a}$ and get that for large $x$ the expansion takes the form
$$
f\left(x\right) \approx  \lim_{b \rightarrow \infty} \left\{f\left(b\right) + \frac{f'\left(b\right)}{x} + \frac{f''\left(b\right)}{2 x^2} + {\cal O}\left(\frac{1}{x^3}\right) \right\} \\
= f\left(\infty\right) + \frac{f'\left(\infty\right)}{x}  + \frac{f''\left(\infty\right)}{2 x^2}  + \dots
$$
So we sort of compute the derivatives at infinity.
I have to stress that I show here an incomplete and simplified explanation to illustrate the idea of how such expansion can be obtained and hope this answers your question. I have not justified exchanging limits, shown convergence, or mentioned the problems with complex branches and various other issues that would need to be addressed in a proper proof.
