How do you find a power series representation for #ln(1-x^2) # and what is the radius of convergence?

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Answer 1 · 2017-05-06T05:06:44

#ln(1-x^2)=sum_(n=0)^oo frac{-(x)^(2n+2)}{n+1}#

The radius of convergence is equal to #1# by the ratio test.

Remember the MacLaurin series representation for #ln(1+u)#:
#color(blue)(ln(1+u)=sum_(n=0)^(oo) frac{(-1)^n(u)^(n+1)}{n+1})#

Substitute #u=-x^2#:

#ln(1-x^2)=sum_(n=0)^(oo) frac{(-1)^n(-x^2)^(n+1)}{n+1}#

#color(white)(ln(1-x^2))= sum_(n=0)^oo frac{(-1)^n(-1)^(n+1)(x)^(2n+2)}{n+1}#

#color(white)(ln(1-x^2))=sum_(n=0)^oo frac{-(x)^(2n+2)}{n+1}#

To find radius of convergence, use ratio test, which states that if #lim_(x->oo) |frac{a_(n+1)}{a_n}|<1#, then #sum_(n=0)^oo a_n# converges

#lim_(n->oo)|frac{color(red)(-)(x)^(2n+4)}{n+2}*frac{n+1}{color(red)(-)(x)^(2n+2)}|#

#=lim_(x->oo) |frac{color(red)((x)^(2n))(x)^4(n+1)}{(n+2)color(red)((x)^(2n))(x)^2}|#

#=|x^2|#

In order for the series to converge, set this less than 1:
#x^2<1, color(red)(x^2> -1)#

The radius of convergence is equal to #1# by the ratio test.