What is the radius of convergence of the series #sum_(n=0)^oo(n*(x+2)^n)/3^(n+1)#?

1 Answer
Oct 20, 2014

Let

#a_n={n(x+2)^n}/{3^{n+1}}#. #Rightarrow a_{n+1}={(n+1)(x+2)^{n+1}}/{3^{n+2}}#

By Ratio Test,

#lim_{n to infty}|{a_{n+1}}/{a_n}|=lim_{n to infty}|{(n+1)(x+2)^{n+1}}/{3^{n+2}}cdot{3^{n+1}}/{n(x+2)^n}|#

by cancelling out common factors,

#=lim_{n to infty}|{(n+1)(x+2)}/{3n}|#

by pulling #{|x+2|}/3# out of the limit and simplifying a bit,

#={|x+2|}/3lim_{n to infty}|1+1/n|={|x+2|}/3<1#

by multiplying by 3,

#Rightarrow |x+2|<3=R#

Hence, the radius of convergence #R# is #3#.


I hope that this was helpful.