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 $R=1/3$