In expressing arctangent as a series, why does substituting $x=1$ make sense? arctangent can be expressed as a power series when $x$ is between $-1$ and $1$.  One post argued that this was possible because the series when x=1 converges, but how do I know it converges to arctangent $1$?
 A: That's a nice question. I suppose that you know that$$\bigl(\forall x\in(-1,1)\bigr):\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^n}{2n+1}.$$But then, by Abel's theorem on power series and since the series $\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$ converges (by the alternate series test), we indeed have\begin{align}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}&=\lim_{x\to1^-}\sum_{n=0}^\infty\frac{(-1)^nx^n}{2n+1}\\&=\lim_{x\to1^-}\arctan x\\&=\arctan 1\\&=\frac\pi4.\end{align}
A: Suppose that $\sum_{n=0}^\infty a_n$ is convergent, to $L$ say. Then
$$f(x)=\sum_{n=0}^\infty a_nx^n $$
converges absolutely whenever $|x|<1$. A theorem of Abel states that
$$L=\lim_{x\to1^-}f(x)=L.$$
A special case is
$$\lim_{x\to1^-}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{2k+1}=
\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}.$$
For $0<x<1$,
$$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}x^{2k+1}
=\sum_{k=0}^\infty\int_0^x(-1)^kt^{2k}
=\int_0^x\sum_{k=0}^\infty(-1)^kt^{2k}=\int_0^x\frac{dt}{1+t^2}
=\arctan x.$$
Here, interchange of sum and integral is justified by uniform convergence.
Therefore,
$$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\lim_{x\to1^-}\arctan x=\frac\pi4.$$
