The Calculus of Infinite Series

We can integrate and differentiate series term by term. So for example,

$$\displaystyle \frac{d}{dx}[\sin(x)] = \displaystyle \frac{d}{dx}\left[\displaystyle \sum^{\infty}_{n=0}(-1)^n \displaystyle \frac{x^{2n+1}}{(2n+1)!}\right]$$

$$= \displaystyle \frac{d}{dx}[x-x^3/3!+x^5/5!-x^7/7!+\dots]$$

$$= 1-x^2/2!+x^4/4!-x^6/6!+\dots$$

$$= \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n}}{(2n)!}$$

We have just found the power series for $f(x)=\cos(x)$ by differentiating each term of the power series for $\sin(x)$.

We can also integrate. The power series for $1/(1+x^2)$ is

$$\displaystyle \frac{1}{1+x^2}=\displaystyle \sum^{\infty}_{n=0}(-1)^nx^{2n}$$

Since $\arctan(x)=\int\displaystyle \frac{1}{1+x^2}\;dx$ we can find its power series by integrating each term of the power series above.

$$\arctan(x) = \displaystyle \int\displaystyle \sum^{\infty}_{n=0}(-1)^nx^{2n}\;dx$$

$$= \displaystyle \sum^{\infty}_{n=0}(-1)^n\displaystyle \frac{x^{2n+1}}{2n+1}$$