Excuse me if I format things wrong; first time posting here. Anyway I am self studying Calc II again and I am just confused on the process of finding the taylor series for $ \ln {1\over (1-x)} $.
My attempt:
I realize $ \ln {1\over (1-x)} $ is the same as $ \ -ln {(1-x)} $. This can then be rewritten as $ \ (-1) \int {1\over (1-x)}dx $. This resembles the geometric series, so we can rewrite it as $ \ (-1) \int \sum_{k=0}^\infty {1\over (1-x)} dx $. We can then switch the position of the integral and sum to get $ \ (-1) \sum_{i=0}^\infty\int x^k dx $. Now here I brought back the $ -1 $ from the outside and reattched it to $ x^k $ so I then got, when integrating, $ \sum_{k=0}^\infty (-1)^k {x^(k+1) \over (k+1)} $. This is just the series for $ ln(1+x) $ (minus the re-indexing) though. But they are two different functions and it's not the correct answer.
I don't care for the answer as I know what it is, I just want to know how to get there. Where did I mess up? And also, is the point of re-indexing when dealing with Taylor series to just get a better form that we can utilize?
TIA
You're steps appear correct but there is an erroneous negative sign thrown in there. You took the known power series for $$ \frac{1}{1-x}=1+x+x^2+x^3+..... $$ And integrated it term by term, which you can do within the radius of convergence of the geometric series and got $$ \int\frac{1}{1-x}dx=\int (1+x+x^2+x^3+.....)dx\\ \Rightarrow -\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+....=\sum_{k=1}^{\infty}\frac{x^k}{k}\\ \ln\frac{1}{1-x}=\sum_{k=1}^{\infty}\frac{x^k}{k} $$