Help with Taylor, ln(1-X), |x|<1

[kevin3295]

Homework Statement

ln(1-X), |x|<1

Homework Equations

Could someone verify if it was developed correctly?

The Attempt at a Solution

$$ln(1-x) = \sum_{n=0}^\infty \left (a_nx^n\right )$$
$$1+a+a^2+a^3+a^4+a^5+a^6... = 1/(1-a)$$
$$a=x$$
$$1+x+x^2+x^3+x^4+x^5+x^6... = 1/(1-x)$$
$$∫1+x+x^2+x^3+x^4+x^5+x^6...dx = ∫1/(1-x) dx$$
$$x+(x^2)/2+(x^3)/3+(x^4)/4+(x^5)/5+(x^6)/6... = - ln(x-1)$$
$$ln(1-x) = - \sum_{n=1}^\infty \left (x^n / n \right )$$

Thanks for your time

 

[Buzz Bloom]

x+(x2)/2+(x3)/3+(x4)/4+(x5)/5+(x6)/6...=ln(x−1)

Hi Keven:

Two issues. One is about your notation of the integrals. There should be a dx.
More important, you have a problem at the quoted step. Think again about

∫1/(1-x)dx = ln(x-1).​

What is the value of ln(x-1) when |x|<1?

Hope this helps.

Regards,
Buzz

 

[kevin3295]

@Buzz Bloom I made the corrections, thank you

 

[Buzz Bloom]

Hi kevin:

Thank you for your thanks.

You still have a problem with -ln(x-1). Think about the value for |x|<1.

Regards,
Buzz