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

1 Answer
Cesareo R.
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

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