Is $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ a power series?

Question

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.

Answer 1

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 $x\in(0,1)$.

Answer 2

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:

1. The series converges only for $x=0$
2. The series converges absolutely for all $x=x_{0}$
3. 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.