In expressing arctangent as a series, why does substituting $x=1$ make sense?

Question

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 $1$?

Answer 1

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}

Answer 2

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, $$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\lim_{x\to1^-}\arctan x=\frac\pi4.$$