Question

How do you find a power series representation for `(x-2)^n/(n^2)` and what is the radius of convergence?

Answer 1

`x in (1,3)` and `sum_{i=1}^{infty}(x-2)^i/i^2 = "PolyLog"(2,x-2)`

Explanation:

Suppose that our quest is for `sum_{i=1}^{infty}(x-2)^i/i^2`.
In this case a variable change `y = (x-2)` give
`sigma(y)=sum_{i=1}^{infty}y^i/i^2`.
In the next steps we will try to build such a power series.
We begin with `sigma_1(y)= sum_{i=0}^{infty}y^n`.
This power series is convergent for `abs(y) < 1` and in this interval is equivalent to

`sigma_1^a(y)= 1/(1-y)`

now integrating `sigma_1(y)` we get

`f(y) = int sigma_1(y)dy = sum_{i=1}^{infty}y^i/i = y sum_{i=1}^{infty}y^{i-1}/i = y sigma_2(y)`

`sigma_2(y)=sum_{i=1}^{infty}y^{i-1}/i` is equivalent to

`sigma_2^a(y)=(int sigma_1^a(y)dy)/y = -(log_e(1-y))/y`

The last step is covered by integrating `sigma_2(y)`.

`int sigma_2(y)dy = sum_{i=1}^{infty}y^i/i^2 = sigma(y)`

or equivalently

`sigma^a(y) = int sigma_2^a(y)dy = int -(log_e(1-y))/y dy= "PolyLog"(2,y)`

The description of function PolyLog can be found in

https://en.wikipedia.org/wiki/Polylogarithm

The convergence interval in `x` is a consequence of `y` interval and is `x in (1,3)`.

The attached figure shows the agreement between `sigma(y)` and `sigma^a(y)`