Represent the following as a power series and state the radius of convergence: $$\ln (\frac{1-x}{1+x})$$
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So far I have that $$\ln(1-x)-\ln(1+x)=\int\limits \frac{-1}{1-x}dx-\int\limits\frac{1}{1+x}dx$$
$$=\sum_{n=0}^{\infty}-x^n-\sum_{n=0}^{\infty}(-1)^nx^n$$
$$=\sum_{n=0}^{\infty}(-x^n-(-1)^nx^n)$$
I'm not sure where to go from here to make it look like a power series?

Question

Represent the following as a power series and state the radius of convergence: $$\ln (\frac{1-x}{1+x})$$
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So far I have that $$\ln(1-x)-\ln(1+x)=\int\limits \frac{-1}{1-x}dx-\int\limits\frac{1}{1+x}dx$$ 
$$=\sum_{n=0}^{\infty}-x^n-\sum_{n=0}^{\infty}(-1)^nx^n$$
$$=\sum_{n=0}^{\infty}(-x^n-(-1)^nx^n)$$
I'm not sure where to go from here to make it look like a power series?

jamesj · Answer

When you integrated the power series you actually didn't; i.e., you haven't integrated them properly.

jamesj · Answer

i.e., it is indeed true that
$$ \frac{1}{1-x} = \sum_{i=0}^{\infty} x^i $$
but you haven't integrated this, term by term.

jamesj · Answer

To be more precise, I should say that expression is true if and only if |x| < 1

jamesj · Answer

However, you don't need to use the integral of the sum to find the power series and honestly you probably shouldn't, as moving infinite sums in and out of integration is fraught with technicalities.  

Just differentiate ln(1-x) and ln(1+x) explicitly.

kirbykirby · Answer

How would I get back my original function by doing so?

jamesj · Answer

I mean differentiate ln(1-x) etc. in order to find the power series, as the coefficients of the power series are the derivatives evaluated at in this case x = 0 divided by 1/n!