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.

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 $x\in(0,1)$.
