# How do you find a power series representation for (arctan(x))/(x)  and what is the radius of convergence?

Dec 31, 2015

Integrate the power series of the derivative of $\arctan \left(x\right)$ then divide by $x$.

#### Explanation:

We know the power series representation of $\frac{1}{1 - x} = {\sum}_{n} {x}^{n} \forall x$ such that $\left\mid x \right\mid < 1$. So $\frac{1}{1 + {x}^{2}} = \left(\arctan \left(x\right)\right) ' = {\sum}_{n} {\left(- 1\right)}^{n} {x}^{2 n}$.

So the power series of $\arctan \left(x\right)$ is $\int {\sum}_{n} {\left(- 1\right)}^{n} {x}^{2 n} \mathrm{dx} = {\sum}_{n} \int {\left(- 1\right)}^{n} {x}^{2 n} \mathrm{dx} = {\sum}_{n} \frac{{\left(- 1\right)}^{n}}{2 n + 1} {x}^{2 n + 1}$.

You divide it by $x$, you find out that the power series of $\arctan \frac{x}{x}$ is ${\sum}_{n} \frac{{\left(- 1\right)}^{n}}{2 n + 1} {x}^{2 n}$. Let's say ${u}_{n} = \frac{{\left(- 1\right)}^{n}}{2 n + 1} {x}^{2 n}$

In order to find the radius of convergence of this power series, we evaluate lim_(n -> +oo)abs((u_(n+1))/u_n.

$\frac{{u}_{n + 1}}{u} _ n = {\left(- 1\right)}^{n + 1} \cdot {x}^{2 n + 2} / \left(2 n + 3\right) \frac{2 n + 1}{{\left(- 1\right)}^{n} {x}^{2 n}} = - \frac{2 n + 1}{2 n + 3} {x}^{2}$.

${\lim}_{n \to + \infty} \left\mid \frac{{u}_{n + 1}}{u} _ n \right\mid = \left\mid {x}^{2} \right\mid$. So if we want the power series to converge, we need $\left\mid {x}^{2} \right\mid = {\left\mid x \right\mid}^{2} < 1$, so the series will converge if $\left\mid x \right\mid < 1$, which is not surprising since it's the radius of convergence of the power series representation of $\arctan \left(x\right)$.