Question

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

Answer 1

Integrate the power series of the derivative of `arctan(x)` then divide by `x`.

Explanation:

We know the power series representation of `1/(1-x) = sum_nx^n AAx` such that `absx < 1`. So `1/(1+x^2) = (arctan(x))' = sum_n (-1)^nx^(2n)`.

So the power series of `arctan(x)` is `intsum_n (-1)^nx^(2n)dx = sum_n int(-1)^nx^(2n)dx = sum_n((-1)^n)/(2n+1)x^(2n+1)`.

You divide it by `x`, you find out that the power series of `arctan(x)/x` is `sum_n((-1)^n)/(2n+1)x^(2n)`. Let's say `u_n = ((-1)^n)/(2n+1)x^(2n)`

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

`(u_(n+1))/u_n = (-1)^(n+1)*x^(2n+2)/(2n+3)(2n+1)/((-1)^nx^(2n)) = -(2n+1)/(2n+3)x^2`.

`lim_(n -> +oo)abs((u_(n+1))/u_n) = abs(x^2)`. So if we want the power series to converge, we need `abs(x^2) = absx^2 < 1`, so the series will converge if `absx <1`, which is not surprising since it's the radius of convergence of the power series representation of `arctan(x)`.