Question 8.6.6: Differentiating and Integrating a Power Series Use the power...

Notice that \sum_{k=0}^{\infty}(-1)^k x^k=\sum_{k=0}^{\infty}(-x)^k is a geometric series with ratio r = −x.

This series converges, then, whenever |r| = |−x| = |x| < 1, to

\frac{a}{1-r}=\frac{1}{1-(-x)}=\frac{1}{1+x}.

That is, for −1 < x < 1, \frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^k x^k. (6.2)

Differentiating both sides of (6.2), we get

\frac{-1}{(1+x)^2}=\sum_{k=0}^{\infty}(-1)^k k x^{k-1}, \quad \text { for }-1<x<1.

Multiplying both sides by −1 gives us a new power series representation:

valid for −1 < x < 1. Notice that we can also obtain a new power series from (6.2) by substitution. For instance, if we replace x with x², we get

\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^k x^k. (6.2)

\frac{1}{1+x^2}=\sum_{k=0}^{\infty}(-1)^k\left(x^2\right)^k=\sum_{k=0}^{\infty}(-1)^k x^{2 k}, (6.3)

valid for −1 < x² < 1 (which is equivalent to having x² < 1 or −1 < x < 1).
Integrating both sides of (6.3) gives us

\int \frac{1}{1+x^2} d x=\sum_{k=0}^{\infty}(-1)^k \int x^{2 k} d x=\sum_{k=0}^{\infty} \frac{(-1)^k x^{2 k+1}}{2 k+1}+c . (6.4)

You should recognize the integral on the left-hand side of (6.4) as \tan ^{-1} x. That is,

\tan ^{-1} x=\sum_{k=0}^{\infty} \frac{(-1)^k x^{2 k+1}}{2 k+1}+c, \quad \text { for }-1<x<1. (6.5)

Taking x = 0 gives us

so that c=\tan ^{-1} 0=0 \text {. }. Equation (6.5) now gives us a power series representation for  \tan ^{-1}x, namely:

\tan ^{-1} x=\sum_{k=0}^{\infty} \frac{(-1)^k x^{2 k+1}}{2 k+1}=x-\frac{1}{3} x^3+\frac{1}{5} x^5-\frac{1}{7} x^7+\cdots, \quad \text { for }-1<x<1.

In this case, the series also converges at the endpoint x = 1.