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.)

In this video, this work is done with a substitution. It is
the same end result, so you can choose your preferred method -- like
the example above or with the substitution.