Power Series and Functions — Question 5

Question 5 – Solution

Step 1: Expand the derivative.

With geometric ratio $-t^2$, $$\frac1{1+t^2}=\sum_{n=0}^{\infty}(-1)^n t^{2n},\qquad |t|<1.$$

Step 2: Integrate from $0$ to $x$.

$$\arctan x=\int_0^x\frac{dt}{1+t^2}=\boxed{\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}},\qquad |x|<1.$$ Integrating the even powers $t^{2n}$ produces only odd powers $x^{2n+1}$; the value at $0$ fixes the constant.

Step 3: Test endpoints.

At $x=1$, the alternating series converges to $\pi/4$. At $x=-1$, it converges to $-\pi/4$. Both are conditional.

Conclusion.

The radius is $R=1$ and the interval is $\boxed{[-1,1]}$.