# A taylor series question

1. May 19, 2009

### kidsmoker

1. The problem statement, all variables and given/known data

(a) Use Taylor's theorem with the Lagrange remainder to show that

$$log(1+x) = \sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k}x^{k}$$

for 0<x<1.

(b) Now apply Taylor's theorem to log(1-x) to show that the above result holds for -1<x<0.

2. Relevant equations

Taylor's theorem w/ Lagrange remainder:

$$f(x) = \sum^{n}_{k=0}\frac{(x-a)^{k}}{k!}f^{(k)}(a) + R(n,x)$$

where

$$R(n,x) = \frac{x^{n+1}}{(n+1)!}f^{(n+1)}(t)$$

for some t in (0,x).

3. The attempt at a solution

I seem to have done part (a) okay by just writing it all out and then showing that the remainder tends to zero as n tends to infinity.

Then when I do the same thing with log(1-x) I get

$$log(1-x) = \sum^{n}_{k=1}\frac{-1}{k}x^{k} + \frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}$$

for some t in (0,x). But we're considering the case here were 0<x<1, so the denominator in the remainder will be between -1 and 0. Consequently, as n tends to infinity, this part will tend to zero and the fraction as a whole will tend to infinity. I've checked my working a few times and tried fiddling about with different methods, including trying the Cauchy remainder, but I can't seem to show that R->0 as n->inf.

Any help appreciated!

2. May 21, 2009

### Born2bwire

Take the magnitude of

$$\frac{(-1)^{n}x^{n+1}}{(n+1)(t-1)^{n+1}}$$

and work to replace the (t-1) with x. Just keep in mind that x and t are both (0,1).

Last edited: May 21, 2009