Taylor series of arcus tangent

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\frac{1}{1\!+\!x^{2}}=1\!-\!x^{2}\!+\!x^{4}\!-\!x^{6}\!+-\ldots,

the radius of convergence of which is $1$ . The series expansion is valid only on the open interval $-1<x<1$ , because the series apparently diverges for $x=\pm 1$ . The power series may be integrated termwise on its interval of convergence, giving

\int_{0}^{x}\!\frac{dt}{1\!+\!t^{2}}=\operatornamewithlimits{\Big/}_{{\!\!\!0}% }^{{\,\quad x}}\!\arctan{t}=\arctan{x}=x\!-\!\frac{x^{3}}{3}\!+\!\frac{x^{5}}{% 5}\!-+\ldots\qquad(|x|<1).

We can show that this Taylor series of arcus tangent is valid also for the end points $x=\pm 1$ of the interval.

We start from the identical equation

\frac{1}{1\!+\!t^{2}}=1\!-\!t^{2}\!+\!t^{4}\!-+\ldots+\!(-1)^{{n-1}}t^{{2n-2}}% +(-1)^{n}\frac{t^{{2n}}}{1+t^{2}},

which can be verified by performing the division $1\!:\!(1\!+\!t^{2})$ . Integrating both sides from $0$ to an arbitrary $x$ , we obtain

\arctan{x}=\underbrace{x\!-\!\frac{x^{3}}{3}\!+\!\frac{x^{5}}{5}\!-+\ldots+\!(% -1)^{{n-1}}\frac{x^{{2n-1}}}{2n\!-\!1}}_{{S_{{2n-1}}}}+\underbrace{(-1)^{n}% \int_{0}^{x}\frac{t^{{2n}}}{1+t^{2}}dt}_{{R_{{2n-1}}}}.

We estimate $R_{{2n-1}}$ :

|R_{{2n-1}}|=\int_{0}^{{|x|}}\frac{t^{{2n}}}{1+t^{2}}dt\leqq\int_{0}^{{|x|}}t^% {{2n}}\,dt=\frac{|x|^{{2n-1}}}{2n\!-\!1}\,\to\,0\quad\mathrm{as}\,\,n\to\infty

for $x=\pm 1$ . Accordingly, when $x=\pm 1$ , we see that

S_{{2n-1}}=\arctan{x}\!-\!R_{{2n-1}}\,\to\,\arctan{x}

as $n\to\infty$ . This means that

\arctan{(\pm 1)}=\pm\frac{\pi}{4}=\pm\left(1\!-\!\frac{1}{3}\!+\!\frac{1}{5}\!% -+\ldots\right).

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