What is the interval of convergence of #sum(2^n(x-3)^n)/(sqrt(n+3))#? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer Eddie Apr 5, 2017 #x in (5/2, 7/2)# Explanation: Ratio test: #lim_(n to 0) abs( ((2^(n+1)(x-3)^(n+1))/(sqrt(n+4)))/((2^n(x-3)^n)/(sqrt(n+3))))# #=lim_(n to 0) abs( (2(x-3)sqrt(n+3))/(sqrt(n+4)))# #=lim_(n to 0) abs( (2(x-3)sqrt(1+3/n))/(sqrt(1+4/n)))# #= 2 abs ((x-3)) lt 1 # Which means either: #- 2 (x-3) lt 1 implies x gt 5/2# # 2 (x-3) lt 1 implies x lt 7/2# #implies x in (5/2, 7/2)# 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 7040 views around the world You can reuse this answer Creative Commons License