How do i solve this problem:
The Geometric series 1/1-x = E(inifinite on the top) (n=0 on the bottom) x^n for |x|

Question

How do i solve this problem:
The Geometric series 1/1-x = E(inifinite on the top) (n=0 on the bottom) x^n for |x|<1 to find the power series and find the interval and radius of convergence of the function f(x)= 1/ (3+2x)

anonymous · Answer

You need to factor once, and substitute.  You know that $$\frac{ 1 }{ 1-x }=\sum_{0}^{\infty}x^n=1+x+x^2+x^3+...$$ with convergence set of $$\left| x \right|<1$$. Since $$\frac{ 1 }{ 3+2x }$$ looks similar, it should have a similar power series representation.  Let's make it look "even more similar" by factoring out a 1/3:$$\frac{ 1 }{ 3 }*\frac{ 1 }{ 1+\frac{ 2x }{3 } }$$ But we need 1 - (something) in the denominator, so rewrite as:$$\frac{ 1 }{ 3}*\frac{ 1 }{ 1-(\frac{ -2x }{ 3 })}$$ Now, replace x in the original series formula and convergence set formula with -2x/3.  Simplify stuff. :)