Show that the power series representation for $\tan ^{-1}x$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \tan ^{-1}x=x-\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}-\dfrac{ x^{7}}{7}+\cdots +(-1)^{n}\dfrac{x^{2n+1}}{2n+1}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{k}x^{2k+1}}{2k+1} }}$$

Find the radius of convergence and the interval of convergence.

Solution If $y=\tan ^{-1}x$, then $y^\prime =\dfrac{1}{1+x^{2}}$, which is the sum of the geometric series $\sum\limits_{k\,=\,0}^{\infty}(-1) ^{k}x^{2k}$. That is, $$\dfrac{1}{1+x^{2}}=\sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}x^{2k}=1-x^{2}+x^{4}-x^{6}+\cdots\,$$

This series converges when $\vert x^{2}\vert < 1$, or equivalently for $-1 < x < 1$. We use the integration property of power series to integrate $y^\prime =\dfrac{1}{1+x^{2}}$ and obtain a series for $y=\tan ^{-1}x$. $$\begin{eqnarray*} \int_{0}^{x}\frac{dt}{1+t^{2}} &=&\int_{0}^{x}(1-t^{2}+t^{4}-\cdots)\, dt \nonumber \\ \tan ^{-1}x &=& x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots +(-1)^{n}\frac{x^{2n+1}}{2n+1}+\cdots = \sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}\dfrac{x^{2k+1}}{2k+1} \end{eqnarray*}$$

The radius of convergence is $1$. To find the interval of convergence, we check $x=-1$ and $x=1$. For $x=-1$, $$-1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots$$

For $x=1$, we get $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$

Both of these series satisfy the two conditions of the Alternating Series Test, and so each one converges. The interval of convergence is the closed interval $[-1,\,1]$.