Hello. For f(x) = loge(1-x), I must first find the MacLaurin series for it, which I did, but then the question continues on to ask me to find the interval of convergence for this MacLauren series, how do I find the interval of convergence? Additional question: I was to find an open interval (e,f) from the previous question, 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 don't understand what the question is asking me to do, could someone help me?

Question

Hello. For f(x) = loge(1-x), I must first find the MacLaurin series for it, which I did, but then the question continues on to ask me to find the interval of convergence for this MacLauren series, how do I find the interval of convergence? Additional question: I was to find an open interval (e,f) from the previous question, 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 don't understand what the question is asking me to do, could someone help me?

experimentx · Answer

http://www.wolframalpha.com/input/?i=expand+ln%281-x%29+at+x%3D0 $$ \left | \frac{x^{n+1} n}{x^n (n+1)}right | < 1 \implies |x| <1$$

experimentx · Answer

$$ \lim_{n \rightarrow \infty }\left | \frac{x^{n+1} n}{x^n (n+1)} \right | < 1 \implies |x| < 1$$

experimentx · Answer

logically i think ...  x<1 ... but i'm getting |x| <1