Question

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

Answer 1

`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)`