How do you find the radius of convergence $Sigma (1-1/n)x^n$ from $n=[1,oo)$ ?

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Answer 1 · 2017-06-10T20:58:26

The radius of convergence is $1$

First note that:

$sum_(n=1)^oo (1-1/n)x^n = sum_(n=2)^oo (1-1/n)x^n$

since the coefficient of the first term is $0$ .

Note also that for $n >= 2$ we have $1/2 <= (1-1/n) < 1$

Hence:

$1/2sum_(n=2)^oo x^n <= sum_(n=2)^oo (1-1/n) x^n <= sum_(n=2)^oo x^n$

Then:

$sum_(n=2)^oo x^n$

is a geometric series with common ratio $x$ , which converges if and only if $abs(x) < 1$

Hence the radius of convergence of the given sum is also $1$ .

How do you find the radius of convergence Sigma (1-1/n)x^nSigma (1-1/n)x^n from n=[1,oo)n=[1,oo)?