Taylor series of $\arctan(x+2)$ at $x=\infty$ The simple question is: what is the correct way to calculate the series expansion of $\arctan(x+2)$ at $x=\infty$ without strange (and maybe wrong) tricks?
Read further only if you want more details.
The first (obvious) problem is that infinite is not a number, thus $\sum_{k=0}^{\infty}\frac{f^{(k)}(\infty)}{k!}(x-\infty)^k$ doesn't make sense.
So I've tried to solve $\lim_{x_0\to \infty} ({\sum_{k=0}^{\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k})$ one term at a time. The first term (with $k=0$) can be easily calculated and the second term is:
$$\lim_{x_0\to \infty} (\frac{x-x_0}{1+(x_0+4x_0+4)} \sim -\frac{x_0}{x_0^2}=\frac{-1}{x_0}) = 0$$
The asymptotic function $\frac{-1}{x_0}$ is similar to what I should get as the second term (which is $\frac{-1}{x}$). But to get the correct result I should do some dangerous stuff: I should replace $x_0$ with $x$ (why?) and do not calculate the limit (otherwise I cannot get the desired precision of the series expansion).
I've also tried to solve $\arctan(1/t+2)$ with $t=0$, but the argument of arctg is still an infinite and the difficulties I encounter are the same.
Is there any way to calculate the series expansion of $\arctan(x+2)$ at $x=\infty$ in a cleaner way without all these problems?
 A: Substituting $x=\frac 1t$ and expanding around $t=0$ is the correct thing to do.  The first derivative of $\tan^{-1}(\frac{1}{t}+2)$ is, by the chain rule
$$
-\frac{1}{t^2}\frac{1}{(\frac 1t +2)^2} = -\frac{1}{(1+2t)^2}~,
$$
which is perfectly finite at $t=0$.  The same goes for all the other derivatives.
After you construct your Taylor expansion, just replace $t$ with $\frac 1x$.

Edit: Perhaps the zeroth-order term concerned you.  Although the argument does indeed become singular, $\tan^{-1}(1/t)$ is well-behaved as $t\to 0^+$; it just approaches $\pi/2$.  So this is the zeroth-order term for your expansion.

Edit 2:
Marc Palm correctly points out in his short answer that the Taylor series does not converge (one easy way to see this is that $\tan^{-1}$ approaches different values at $x \to \pm \infty$, but these both correspond to $t\to 0$, so the function cannot be analytic there; this is why I wrote $t\to 0^+$ above).  However, the procedure above still gives you a sensible expansion to some finite order, with error bounded by Taylor's theorem.
A: The poles accumulate at $\infty$, no Taylor expansion at $\infty$ is possible. $artcan$ is a transcendental function and has non-removable singularity at $\infty$.
A: We can write $$\arctan (x+2) = \frac{\pi}{2} - \int_x^\infty \frac{dt}{1+(t+2)^2}.$$
Now we can expand the integrand into a "power series" in $\frac1t$,
$$\begin{align}
\frac{1}{1+(t+2)^2} &= \frac{1}{2i} \left(\frac{1}{t+2-i} - \frac{1}{t+2+i}\right)\\
&= \frac{1}{2it} \left(\frac{1}{1+\frac{2-i}{t}} - \frac{1}{1+\frac{2+i}{t}}\right)\\
&=\sum_{\nu=1}^\infty \frac{(-1)^\nu\bigl((2-i)^\nu - (2+i)^\nu\bigr)}{2i}\frac{1}{t^{\nu+1}}.
\end{align}$$
That series converges uniformly on $[c,\infty)$ for $c > \sqrt{5}$, and all summands as well as the sum are integrable, so
$$\begin{align}
\arctan(x+2) &= \frac{\pi}{2} - \sum_{\nu=1}^\infty \frac{(-1)^\nu\bigl((2-i)^\nu - (2+i)^\nu\bigr)}{2i} \int_x^\infty \frac{dt}{t^{\nu+1}}\\
&= \frac{\pi}{2} - \sum_{\nu=1}^\infty \frac{(-1)^\nu\bigl((2-i)^\nu - (2+i)^\nu\bigr)}{2i\nu} \frac{1}{x^\nu}.
\end{align}$$
Expanding the first few coefficients yields
$$\arctan (x+2) = \frac{\pi}{2} - \frac{1}{x} + \frac{2}{x^2} - \dotsb$$
