Question

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

Answer 1

`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`