MacLaurin series/Ratio test.

**For f(x) = loge(1-x) **

*Clarification: The question isn’t loge(1+x), it’s loge(1-x) *

**a)** Find the MacLaurin series: f(0) +f'(0)x + f''(0)/2! + f'''(0)/3!....etc

My question is, to what extent do I keep applying the series to, since the series goes on forever and there are no constraints specified.

**b)** Also, I’m having trouble finding the interval of convergence for the MacLauren series, how do I find the interval of convergence?

**c)** According to the question I am given, I was to find an open interval (e,f) from b) since the statement of the ratio test uses a strict inequality.

And now I have to use an appropriate test of series convergence for the MacLaurin series found at the start at x = e and x = f

I am clueless as to how to answer these questions, could somebody please show me the steps? I am keen to understand how to go about tackling these questions.

Thank you very much, I really appreciate your help.

Re: MacLaurin series/Ratio test.

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$ for $\displaystyle |x| < 1$

$\displaystyle \int \frac{dx}{1-x} = C - x - \frac{x^2}{2} - \frac{x^3}{3} - ...$

$\displaystyle \ln(1-x) = C - x - \frac{x^2}{2} - \frac{x^3}{3} - ...$

$\displaystyle x = 0 \implies C = 0$

$\displaystyle \ln(1-x) = \left(- x - \frac{x^2}{2} - \frac{x^3}{3} - ... \right) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$

interval of convergence is $\displaystyle -1 \le x < 1$ ,

now ... why is x = -1 included and x = 1 not ?

Re: MacLaurin series/Ratio test.