What is the radius of convergence of #sum_1^oo (x-1)^n / 2^(2n)?#? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer Cesareo R. Jul 9, 2016 #3 < x < 5# Explanation: #sum_1^oo (x-1)^n / 2^(2n)=sum_1^oo ((x-1)/4)^n# Making #y = (x-1)/4# #sum_1^oo (x-1)^n / 2^(2n) equiv sum_1^oo y^n# We know that for #abs y < 1# # sum_1^oo y^n = 1/(1-y)# or equivalently #sum_1^oo (x-1)^n / 2^(2n)=4/(5-x)# for #abs((x-1)/4)<1# or # 3 < x < 5# Answer link Related questions How do you find the radius of convergence of a power series? How do you find the radius of convergence of the binomial power series? What is the radius of convergence for a power series? What is interval of convergence for a Power Series? How do you find the interval of convergence for a power series? How do you find the radius of convergence of #sum_(n=0)^oox^n# ? What is the radius of convergence of the series #sum_(n=0)^oo(x-4)^(2n)/3^n#? How do you find the interval of convergence for a geometric series? What is the interval of convergence of the series #sum_(n=0)^oo((-3)^n*x^n)/sqrt(n+1)#? What is the radius of convergence of the series #sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)#? See all questions in Determining the Radius and Interval of Convergence for a Power Series Impact of this question 1859 views around the world You can reuse this answer Creative Commons License