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#.