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

May 27, 2016

$x \in \left(1 , 3\right)$ and ${\sum}_{i = 1}^{\infty} {\left(x - 2\right)}^{i} / {i}^{2} = \text{PolyLog} \left(2 , x - 2\right)$

#### Explanation:

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

${\sigma}_{1}^{a} \left(y\right) = \frac{1}{1 - y}$

now integrating ${\sigma}_{1} \left(y\right)$ we get

$f \left(y\right) = \int {\sigma}_{1} \left(y\right) \mathrm{dy} = {\sum}_{i = 1}^{\infty} {y}^{i} / i = y {\sum}_{i = 1}^{\infty} {y}^{i - 1} / i = y {\sigma}_{2} \left(y\right)$

${\sigma}_{2} \left(y\right) = {\sum}_{i = 1}^{\infty} {y}^{i - 1} / i$ is equivalent to

${\sigma}_{2}^{a} \left(y\right) = \frac{\int {\sigma}_{1}^{a} \left(y\right) \mathrm{dy}}{y} = - \frac{{\log}_{e} \left(1 - y\right)}{y}$

The last step is covered by integrating ${\sigma}_{2} \left(y\right)$.

$\int {\sigma}_{2} \left(y\right) \mathrm{dy} = {\sum}_{i = 1}^{\infty} {y}^{i} / {i}^{2} = \sigma \left(y\right)$

or equivalently

${\sigma}^{a} \left(y\right) = \int {\sigma}_{2}^{a} \left(y\right) \mathrm{dy} = \int - \frac{{\log}_{e} \left(1 - y\right)}{y} \mathrm{dy} = \text{PolyLog} \left(2 , y\right)$

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 $x \in \left(1 , 3\right)$.

The attached figure shows the agreement between $\sigma \left(y\right)$ and ${\sigma}^{a} \left(y\right)$