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

1 Answer
May 27, 2016

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)

enter image source here