<s>[INST] Is it possible to have a power series for arctan(x) centered at 1? My Calculus 2 professor referenced that such a series is impossible, but why? I understand how to properly find the power series of arctan(x) centered at 0.
 [/INST] Since
$$
\begin{align}
\tan(\arctan(1+x)-\arctan(1))
&=\frac{(1+x)-1}{1+(1+x)\cdot1}\\
&=\frac x{2+x}\tag{1}
\end{align}
$$
we have
$$
\begin{align}
\arctan(1+x)
&=\frac\pi4+\arctan\left(\frac{x}{2+x}\right)\\
&=\frac\pi4+\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac x{2+x}\right)^{2k+1}\tag{2}
\end{align}
$$
Expanding $(2)$ using the binomial theorem, the coefficient of $x^n$, for $n\ge1$, is
$$
\begin{align}
&\frac{\left(-\frac12\right)^n}n\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}(-1)^{k+1}\binom{n}{2k+1}\\
&=\frac{\left(-\frac12\right)^n}n\sum_{k=0}^n\frac{(-i)^k-i^k}{2i}\binom{n}{k}\\
&=\frac{\left(-\frac12\right)^n}n\frac{(1-i)^n-(1+i)^n}{2i}\\
&=\frac{\left(-\frac1{\sqrt2}\right)^n}n\frac{\left(\frac{1-i}{\sqrt2}\right)^n-\left(\frac{1+i}{\sqrt2}\right)^n}{2i}\\
&=-\frac{\left(-\frac1{\sqrt2}\right)^n}n\sin\left(\frac{n\pi}4\right)\tag{3}
\end{align}
$$
Therefore,
$$
\arctan(1+x)=\frac\pi4-\sum_{n=1}^\infty\frac{\sin\left(\frac{n\pi}4\right)}n\left(-\frac{x}{\sqrt2}\right)^n\tag{4}
$$
Note that the series in $(2)$ converges for $-1\lt x\lt\infty$ and $(4)$ matches Jack D'Aurizio's answer.
</s>