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)

Answer 1

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. :)