Power Series Centered at $x=a$.

Power series don't have to be centered at $0$.  They can be centered at any point $x=a$, in which case the series will contain powers of $(x-a)$ instead of powers of $x$.  Since the interval of convergence of a power series has a center that is the $x$-value where the series is centered, if a power series centered at $a$ has radius of convergence $R$, its interval of convergence will be from $a-R$ to $a+R$.  All possible intervals of convergence of such a series are:  $$[a-R,a+R],\quad(a-R,a+R],\quad[a-R,a+R),\quad(a-R,a+R),\quad\{a\}=[a,a]\quad\text{ and }\quad(-\infty,\infty).$$

Example: The power series $$\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}(x-1)^n}{n}}$$ is centered at $a=1$, which you determine when you look at the power of $x$, which is actually a power of $x-1=x-a$.

As before, we can use the Ratio or Root Test for determining the radius of convergence, and the interval of convergence will be centered at $x=1$. 

DO:  find the radius and interval of convergence, just as we have before with power series centered at the origin.

Solution:  $\displaystyle\sqrt[n]{\left|\frac{(x-1)^n}{n}\right|}=\frac{\left|x-1\right|}{\sqrt[n]n}\longrightarrow \big|x-1\big|$.  By the Root Test, this series converges when $\big|x-1\big|<1$, so $R=1$.  By algebra, this inequality is the same as $-1<x-1<1$, so $0<x<2$.  Our interval of convergence is the interval from $0$ to $2$.  You see that it is centered at $x=1$, which is where our series is centered.  Now we check our endpoints.  

When $x=0$, we have $\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}(x-1)^n}{n}}=\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}(0-1)^n}{n}}=\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{2n+1}}{n}}=\displaystyle{\sum_{n=1}^\infty -\frac{1}{n}}$ which is the divergent (negative) harmonic series.

When $x=2$, we have $\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}(x-1)^n}{n}}=\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}(2-1)^n}{n}}=\displaystyle{\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}}=$ which is the convergent alternating harmonic series.  

$R=1$, and the interval of convergence of $(0,2]$. (This series adds up to $\ln(x)$, by the way.)