A series of the form $\sum _{n\=0}^{\infty }{a}_n{\left(x-c\right)}^n$ is called a power series centered at $c$. The behavior of such power series is exactly parallel to that of the series centered at $c\=0$.

The convergence of a power series is determined pointwise: the variable $x$ is replaced throughout with a constant $x_0$ and the convergence of the resulting series of numbers, namely $\mathrm{\Σ} a_n x_0^n$, is determined as per Section 8.3. If this series of numbers converges, then the series is said to converge for $x\=x_0$. The totality of $x$-values for which the power series converges forms the domain of a function $f\left(x\right)$ whose rule is the power series.

The repetition of $\mathrm{\&Sigma;} a_n{x}^n$ in item (5) in the table is deliberate. The radius of convergence for this series can be established by the Ratio test as follows. The series converges if, by the Ratio test, $\underset{n\to \infty }{lim}\left|\frac{a_{n\&plus;1} x^{n\&plus;1}}{a_n x^n}\right|<1$. For fixed $x$, routine algebra gives $\left|x\right|$$<\underset{n\to \infty }{lim}\&verbar;\frac{a_n}{a_{n\&plus;1}}\&verbar;\equiv R$.

Consider instead, the series $\mathrm{\&Sigma;} a_n x^{2 n}$ and apply the same reasoning. The Ratio test now gives

The radius of convergence for a power series in powers of $\mathrm{\&alpha;} x\&plus;\mathrm{\&beta;}$ must take into account the factor $\left|\mathrm{\&alpha;}\right|$. Indeed, the radius of convergence will then be  $\frac{1}{\left|\mathrm{\&alpha;}\right|}\underset{n\to \infty }{lim}\&verbar;\frac{a_n}{a_{n\&plus;1}}\&verbar;$ because the calculations that lead to $R$ in the simpler case will lead to $\left|\mathrm{\&alpha;} x\&plus;\mathrm{\&beta;}\left|<\underset{n\to \infty }{lim}\right|a_n\&sol;a_{n\&plus;1}\right|$. If the limit on the right is taken as $\mathrm{\&lambda;}$, the inequality then implies $\left|x\&plus;\mathrm{\&beta;}\&sol;\mathrm{\&alpha;}\right|<\mathrm{\&lambda;}\&sol;\left|\mathrm{\alpha }\right|$.

The term $\mathrm{\&beta;}\&sol;\mathrm{\&alpha;}$ shifts the center of the interval of convergence to $x\&equals;-\mathrm{\&beta;}\&sol;\mathrm{\&alpha;}$, but the factor $1\&sol;\left|\mathrm{\&alpha;}\right|$ scales $\mathrm{\&lambda;}$.

The endpoints of the interval of convergence are then $x\&equals;-\mathrm{\&beta;}\&sol;\mathrm{\&alpha;}-\mathrm{\&lambda;}\&sol;\left|\mathrm{\&alpha;}\right|$ and $x\&equals;-\mathrm{\&beta;}\&sol;\mathrm{\&alpha;}\&plus;\mathrm{\&lambda;}\&sol;\left|\mathrm{\alpha }\right|$.