How i calculate the value of the sum #2^(n+3)/(n!)# ? Calculus Power Series Determining the Radius and Interval of Convergence for a Power Series 1 Answer Cesareo R. Jun 18, 2016 # 2^3e^2# Explanation: #e^x = sum_{i=0}^{oo}x^i/(i!) # so #e^2 = sum_{i=0}^{oo}2^i/(i!) # then #2^3sum_{i=0}^{oo}2^i/(i!) = sum_{i=0}^{oo}2^{i+3}/(i!) = 2^3e^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 1737 views around the world You can reuse this answer Creative Commons License