Consider the series:
#sum_(n=0)^oo (x-3)^n/4^(n+1)= 1/4sum_(n=0)^oo ((x-3)/4)^n#
This is a geometric series of ratio #(x-3)/4#, so it is absolutely convergent for:
#abs ((x-3)/4) < 1#
that is for #x in (-1,7)#, and in this interval its sum is:
#1/4sum_(n=0)^oo ((x-3)/4)^n = 1/4 1/(1-(x-3)/4) = 1/(4-x+3) = 1/(7-x)#
As the series is absolutely convergent we can integrate it term by term and the resulting series will have the same interval of convergence:
#1/4sum_(n=0)^oo int_3^x ((t-3)/4)^ndt = int_3^x 1/(7-t)dt#
Clearly:
#1/4 int_3^x((t-3)/4)^ndt = int_3^x((t-3)/4)^nd((t-3)/4) = 1/(n+1) ((x-3)/4)^(n+1)#
So:
#sum_(n=0)^oo 1/(n+1) ((x-3)/4)^(n+1) = int_3^x 1/(7-t)dt =[-ln abs(7-t)]_3^x = ln4 -ln abs (7-x) = ln abs (4/(7-x))#
Since the series is convergent for #x in (-1, 7)# the argument is always positive and we can omit the absolute value:
#sum_(n=0)^oo 1/(n+1) ((x-3)/4)^(n+1) = ln (4/(7-x))#
