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