What is the interval of convergence of $sum_1^oo (2^n(n+1))/(x-1)^n$ ?

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Answer 1 · 2016-07-24T22:46:20

$x < -1$ and $x > 3$

$sum_1^oo (2^n(n+1))/(x-1)^n = sum_1^oo (n+1)((2)/(x-1))^n$

but

$d/(dx)sum_1^oo (2/(x-1))^{n+1} =-2/(x-1)^2 sum_1^oo(n+1) (2/(x-1))^n$

then

$sum_1^oo(n+1) (2/(x-1))^n = -(x-1)^2/2d/dx(sum_1^oo (2/(x-1))^{n+1} )$

but for $abs(2/(x-1))<1$ we have

$sum_1^oo (2/(x-1))^{n+1} = 1/(1-2/(x-1))-1 = (x-1)/(x-3)-1$

and

$d/dx(sum_1^oo (2/(x-1))^{n+1} ) = 2/(x-3)^2$

Finally

$sum_1^oo(n+1) (2/(x-1))^n = ((x-1)/(x-3))^2-1$

This sum is convergent for $abs(2/(x-1))<1$ or

$x < -1$ and $x > 3$

What is the interval of convergence of sum_1^oo (2^n(n+1))/(x-1)^n sum_1^oo (2^n(n+1))/(x-1)^n ?