# Power series and integration >

1. Mar 31, 2007

### FaNgS

Power series and integration >:(

Hello, this question is about power series and integration of power series, the question and my working is on the image below.
I had to write the question and my working plus the correct answer on a piece of paper and scan it, sorry for the hasle

http://i148.photobucket.com/albums/s38/InsertMe/img015.jpg

Smilies in the image :P

2. Mar 31, 2007

### AKG

$$-\frac{1}{x} - \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n-1}}{4n^2-1}$$

$$= -\frac{1}{x} + \sum_{n=0}^{\infty}\frac{-(-1)^nx^{2n-1}}{4n^2-1}$$

$$= -\frac{1}{x} + \sum_{n=0}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}$$

$$= -\frac{1}{x} + \frac{(-1)^{0+1}x^{2\cdot 0-1}}{4\cdot 0^2-1} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}$$

$$= -\frac{1}{x} + \frac{x^{-1}}{1} + \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}$$

$$=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{4n^2-1}$$

3. Apr 1, 2007

### FaNgS

thanks alot AKG

just one question is it always possible to do that whenever i get a function plus (or minus) a power series?

4. Apr 1, 2007

### HallsofIvy

Staff Emeritus
In this particular case you should have noticed that when n= 0, x2n-1 is just x-1, the same as the 1/x you are adding. Strictly speaking, what you have is not a "power series" since the standard definition requires non-negative powers only.