Is $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ a power series? The series $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ can be expressed as the following geometric series: 
$$\sum_{n=0}^\infty \left(\frac{x^2-5}{2}\right)^n.$$ 
This series should converge if $$\left|\frac{x^2-5}{2}\right|<1.$$
This gives the possible values of $x$ to lie in the range $$(-\sqrt7,-\sqrt3) \cup (\sqrt3,\sqrt7).$$ This would mean that the above series is not a power series as a power series cannot have any discontinuities in its interval of convergence. However the series can be rewritten in the following form: $$\sum_{n=0}^\infty \frac{((x-\sqrt5)(x+\sqrt5))^n}{2^n}$$ 
which can further be rewritten as 
$$\sum_{n=0}^\infty \frac{((x-\sqrt5)^2+2\sqrt5(x-\sqrt5))^n}{2^n}.$$
This final series can be rewritten in the form 
$$\sum_{n=0}^\infty a_n(x-\sqrt5)^n$$ 
by using the binomial expansion. According to the definition, a power series is any series of the form $$\sum_{n=0}^\infty a_n(x-c)^n.$$ 
This would mean that the above series is a power series with center $\sqrt5$. Thus there appears to be a contradiction which I am not able to resolve.
 A: If that series converges, then its sum will be equal to the sum of the original series on some interval centered at $\sqrt5$, yes. But its region of convergence might be, say, $\left(\sqrt3,2\sqrt5-\sqrt3\right)$.
Here's a similar situation: $\sum_{n=0}^\infty x^n=\frac1{1-x}$ when $x\in(-1,1)$. But if you write $x$ as $\left(x-\frac12\right)+\frac12$ and you expand this, then you will get a power series centered at $\frac12$ which also converges to $\frac1{1-x}$, but only when $x\in(0,1)$.
A: A power series is any series of the form
$$\sum_{n} a_{n} \, (x-b)^n$$
which leads to saying the series in question is not a power series. 
The two forms 
$$\sum_{n=0}^{\infty} \left(\frac{x^2 - a^2}{b}\right)^n \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{b^n} \, ( (x - a)^2 + 2 a \, (x-a) )^n$$
provide the same result seen as follows:
$$ \sum_{n=0}^{\infty} \left(\frac{x^2 - a^2}{b}\right)^n = \frac{b}{b + a^2 - x^2}$$
and
\begin{align}
\sum_{n=0}^{\infty} \frac{1}{b^n} \, ( (x - a)^2 + 2 a \, (x-a) )^n &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n}{k} \, \frac{(2 a)^k}{b^n} \, (x-a)^{2n-k} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{n+k}{k} \, \frac{(2 a)^k}{b^{n+k}} \, (x-a)^{2n+k} \\
&= \sum_{n=0}^{\infty} \frac{(x-a)^{2n}}{b^n} \, \sum_{k=0}^{\infty} \binom{n+k}{k} \, \frac{(2 a (x - a))^k}{b^k} \\
&= \sum_{n=0}^{\infty} \frac{(x-a)^{2n}}{b^n} \, \frac{b^{n+1}}{(b + 2 a^2 - 2 a x)^{n+1}} \\
&= \frac{b}{b + 2 a^2 - 2 a x} \, \sum_{n=0}^{\infty} \left(\frac{(x-a)^2}{b + 2 a^2 - 2 a x}\right)^n \\
&= \frac{b}{b + a^2 - x^2}.
\end{align}
This is an indicator that both forms do not fit the definition of a power series.
This form brings into question the definitions of double sums where the coefficients are not strictly constants.
In general:
For any power series, one of the following is true:


*

*The series converges only for $x=0$

*The series converges absolutely for all $x=x_{0}$

*The series converges absolutely for all $x$ in some finite open interval $(-R,R)$ and diverges if $x<-R$ or $x>R$. At the points $x=R$ and $x=-R$, the series may converge absolutely, converge conditionally, or diverge. 


Notes
Use was made of
\begin{align}
\sum_{n=0}^{\infty} \sum_{k=0}^{n} B(n,k) &= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} B(n+k,k) \\
\sum_{k=0}^{\infty} \binom{n+k}{k} \, t^k &= \frac{1}{(1-t)^{n+1}}
\end{align}
to demonstrate the second series equals the first series when evaluated.
