What is the interval of convergence of #sum_1^oo ln(n)/e^n (x-e)^n #? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer Cesareo R. Jan 28, 2017 #abs((x-e)/e)<1# Explanation: #((x-e)/e)^n < log(n)((x-e)/e)^n <(n+1)((x-e)/e)^n=ed/(dx)((x-e)/e)^(n+1) # so the series is convergent for #abs((x-e)/e)<1# 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 1565 views around the world You can reuse this answer Creative Commons License