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#