What is the interval of convergence of #sum_2^oo (x+1)^n /( n^2 -2n+1) #?

1 Answer
Jun 16, 2016

#S(x) =- (x+1) (-1.64493+ Log_e(-x) Log_e(1 + x) + "PolyLog"(2, -x))#

with #-2 < x < 0#

Explanation:

Considering the general term of #S(x)=sum_{i=2}^oo (x+1)^i /(i-1)^2# as
#t_i = (x+1)^i/(i-1)^2#
we construct the following relations
#d/(dx)(t_i/(x+1))=((x+1)^{i-2})/(i-1)#
#d/(dx)((x+1)d/(dx)(t_i/(x+1)))=(x+1)^{i-2}#
#(x+1)^2d/(dx)((x+1)d/(dx)(t_i/(x+1)))=(x+1)^i#
but
#sum_{i=2}^{oo}(x+1)^i =- (1/x+2+x)# for #-2 < x < 0#
then
#sum_i d/(dx)((x+1)d/(dx)(t_i/(x+1)))=-1/(x+1)^2 (1/x+2+x)#
and
#sum_id/(dx)(t_i/(x+1)) =-1/(x+1) int 1/(x+1)^2 (1/x+2+x) dx#

proceeding this way we get

#S(x)=sum_i t_i=-(x+1) (c_2 + c_1 Log_e(1 + x) + Log_e(-x)Log_e(1 + x) + "PolyLog"(2, -x))#

Attached the comparisson between #sum_{i=2}^oo (x+1)^i /(i-1)^2# and #S(x)#

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