What is the interval of convergence of #sum_1^oo [(3x)^n(x-2)^n]/(nx) #?

1 Answer
Jul 1, 2016

#sum_1^oo [(3x)^n(x-2)^n]/(nx)=-(LogAbs[3 (x-2) x-1]]/x# and converges for #x ne 0#
#1/3 (3 - 2 sqrt[3]) < x < 1/3 (3 - sqrt[6])#
#1/3 (3 + sqrt[6]) < x < 1/3 (3 + 2 sqrt[3])#

Explanation:

#sum_1^oo [(3x)^n(x-2)^n]/(nx) = 1/x sum_1^oo [(3x)^n(x-2)^n]/n#

but

#d/(dx)( ((3x)^n(x-2)^n)/n) = 6 (x - 1) ((3 x) (x - 2))^(k - 1)#

or

#d/(dx)(sum_1^oo [(3x)^n(x-2)^n]/n)=6(x-1)sum_1^{oo}((3 x) (x - 2))^(k - 1)#

Now supposing that

#abs(3 x (x - 2))<1#

#sum_1^{oo}((3 x) (x - 2))^(k - 1) = -1/(3 x (x - 2) - 1)#

then

#d/(dx)(sum_1^oo [(3x)^n(x-2)^n]/n) = -(6 (x - 1))/(3 x (x - 2) - 1)#

Integrating again and dividing by #x# we obtain

#sum_1^oo [(3x)^n(x-2)^n]/(nx) =-(LogAbs[3 (x-2) x-1]]/x #