MacLaurin Series and Ratio Test for f(x) = loge(1-x)

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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 attempt 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.
 
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You could start by trying to find f(0), f'(0), f''(0) etc. Any ideas on that? Can you find a general way to write the nth derivative of f evaluated at 0?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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