Radius of Convergence for Power Expansion?

Question

Expanded as a power series in terms of #x - 2/3#, what is the radius of convergence of #f(x) = 1/(x(1-x))#?

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Answer 1 · 2017-02-28T13:20:24

#1/(x(1-x))= -sum_(n=0)^oo (1+(-1/2)^(n+1))3^(n+1)(x-2/3)^n#

with radius of convergence #R=1/3#

We have:

#f(x) = 1/(x(1-x)) = 1/x + 1/(1-x)#

Substitute: #t = (x-2/3)#

#f(t) = 1/(t+2/3)+ 1/(1-(t+2/3)) = 1/(2/3+t) + 1/(1/3-t) = 3/2 1/(1+3/2t)- 3/(1-3t)#

Both terms can be expressed as sum of a geometric series with ratio: #(-(3t)/2)# and #(3t)# respectively:

#1/(1+(3t)/2) = sum_(n=0)^oo ((-3t)/2)^n# for #abs ((3t)/2) < 1 => t in (-2/3,2/3)#

#1/(1-3t) = sum_(n=0)^oo (3t)^n# for #abs (3t) < 1 => t in (-1/3,1/3)#

So we have that in the intersection of the two intervals, that is for #t in (-1/3,1/3)#

#f(t) = 3/2sum_(n=0)^oo ((-3t)/2)^n-3 sum_(n=0)^oo (3t)^n#

#f(t) = -sum_(n=0)^oo (-3/2)^(n+1)t^n- sum_(n=0)^oo 3^(n+1)t^n#

#f(t) = -sum_(n=0)^oo 3^(n+1)t^n (1+(-1/2)^(n+1))#

and substituting back #x#:

#1/(x(1-x))= -sum_(n=0)^oo (1+(-1/2)^(n+1))3^(n+1)(x-2/3)^n#

with radius of convergence #R=1/3#