]> 30.5 Manipulating Absolutely Convergent Series

## 30.5 Manipulating Absolutely Convergent Series

One of the nice features of absolutely convergent series is that you can manipulate and rearrange them to suit your fancy.

Thus, given such a power series, you can integrate term by term or differentiate term by term within its radius of convergence, where it absolutely converges.

$( 1 − x ) − 1 = 1 + x + x 2 + ⋯ + x n + ⋯$

and integrate both sides, integrating each term on the right hand side.

We get

$− ln ⁡ ( 1 − x ) = x + x 2 2 + ⋯ + x n n + ⋯$

Exercise 30.10 Integrate both sides again and use the identity $1 n ( n + 1 ) = 1 n − 1 n + 1$ to rearrange and interpret the right hand side.

We can differentiate the geometric series term by term as well.

Notice that upon integration the terms got slightly more convergent than they were. On differentiation, they get slightly less convergent, but not enough to affect the radius of convergence

$( 1 − x ) − 2 = 1 + 2 x + 3 x 2 + ⋯ + ( n + 1 ) x n + ⋯$

Exercise 30.11 Differentiate both sides here again to find a series expansion for $( 1 − x ) 3$ in the variable $x$ . Check your result for $x = 0$ and $x = 1 10$ .