How do you find the radius of convergence #Sigma (1-1/n)x^n# from #n=[1,oo)#?
1 Answer
Jun 10, 2017
The radius of convergence is
Explanation:
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
Note also that for
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
Hence the radius of convergence of the given sum is also