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.
 A: Impossible is nothing. By termwise integration of the Taylor series of $\frac{1}{1+x^2}$ centered at $x=1$ we easily get the Taylor series of $\arctan(x)$ centered at $x=1$. Its radius of convergence is $\sqrt{2}$ since the closest singularities to $x=1$ lie at $\pm i$.
$$\arctan(x)=\frac{\pi}{4}+\sum_{m\geq 0}\left(\frac{(-1)^m(x-1)^{4m+1}}{2^{2m+1}(4m+1)}-\frac{(-1)^m(x-1)^{4m+2}}{2^{2m+1}(4m+2)}+\frac{(-1)^m(x-1)^{4m+3}}{2^{2m+2}(4m+3)}\right)$$
Since, for instance, $\frac{\pi}{4}=\arctan\left(\frac{1}{2}\right)+\frac{1}{2}\arctan\left(\frac{3}{4}\right)$, that Taylor series can be exploited to provide fast-converging series to $\pi$ of the BBP-type, like the celebrated
$$ \pi = \sum_{k\geq 0}\frac{1}{16^k}\left(\frac{4}{8k+1}-\frac{2}{8k+4}-\frac{1}{8k+5}-\frac{1}{8k+6}\right).$$
A: 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.
