$a \in \R$. A formal power series centered at $a $ is any series of the form
$$\sum_{ n = 0 }^{ \infty } c_n (x - a)^n $$
where $c_1, _2, … $ is a sequence of real number (not depending on $x $); we refer to $c_n $ as the $n$th coefficient of this series.
Note that each term $c_n(x - a)^n $ in this series is a function of a real variable $x$.
The series $\sum_{ n=0 }^{ \infty } n! (x - 2)^n $ is a formal power series centered at $2$. The series $\sum_{ n=0 }^{ \infty } 2^x (x - 3)^n $ is not a formal power series since the coefficients $2^x $ depend on $x$.
Let $\sum_{ n=0 }^{ \infty } c_n (x - a)^n $ be a formal power series. We define the radius of convergence $R $ of this series to be the quantity
$$R:= \frac{ 1 }{ \limsupu{ n }{ \infty } \abs{ c_n }^{1 / n}} $$
where we adopt the convention that
$$\begin{align*} \frac{ 1 }{ 0 } &= + \infty \br \frac{ 1 }{ + \infty } &= 0 \end{align*}$$
Notice that we are discussing this in $\R \cup { +\infty, -\infty } $
Each number $\abs{ c_n }^{ 1 / n } $ is non-negative, so the limit $\limsupu{ n }{ \infty } \abs{ c_n }^{ \frac{ 1 }{ n }} $ can take on any value from $0 $ to $+ \infty $, inclusive. Thus $R$ can also take on any value between $0 $ and $+ \infty $ inclusive.
Notice that radius of convergence always exists, even if the sequence $\abs{ c_n }^{ 1/ n } $ is not convergent, because the lim sup of any sequence always exists in $\R \cup { +\infty, -\infty } $.
The series $\sum_{ n=0 }^{ \infty } n(-2)^n (x - 3)^n $ has radius of convergence
$$\limu{ 1 }{ \limsupu{ n }{ \infty } \abs{ n (-2)^n }^{ 1/ n }} = \frac{ 1 }{ \limsupu{ n }{ \infty } 2n^{ 1/ n }} = \frac{ 1 }{ 2 } $$
The series $\sum_{ n=0 }^{ \infty } 2^{n^2} (x + 2) ^n$ has radius of convergence
$$\frac{ 1 }{ \limsupu{ n }{ \infty } \abs{ 2^{-n^2}}^{ 1/ n }} = \frac{ 1 }{ \limsupu{ n }{ \infty } 2^{-n}} = \frac{ 1 }{ 0 } = + \infty $$
Let $\sum_{ n=0 }^{ \infty } c_n (x - a)^n $ be a formal power series, and let $R $ be its radius of convergence.
- (Divergence outside of the radius of convergence) If $x \in \R $ is such that $\abs{ x - a } > R $, then the series $\sum_{ n=0 }^{ \infty } c_n (x- a)^n $ is divergent for that value of $x $.
- (Convergence inside the radius of convergence) If $x \in \R $ is such that
$\abs{ x - a } < R $, then the series $\sum_{ n=0 }^{ \infty } c_n (x - a)^n $ is absolutely convergent for that value of $x $.
For parts 3. - 5., assume that $R > 0 $(i.e. the series converges at at least one other point than x = a). Let $f: (a - R, a + R) \to \R $ be the function $f(x) := \sum_{ n = 0 }^{ \infty } c_n (x -a)^n$; this function is guaranteed to exist by 2.. - (Uniform convergence on compact sets) For any $0 < r < R $, the series $\sum_{ n=0 }^{ \infty } c_n (x- a)^n $ converges uniformly to $f $ on the compact interval $[a- r, a +r] $. In particular, $f $ is continuous on $(a - R, a + R) $.
- (Differentiation of power series) The function $f $ is differentiable on $(a -R, a + R)$, and for any $0 < r <R $, the series $\sum_{ n=1 }^{ \infty }n c_n(x - a)^{n - 1} $ converges uniformly to $f' $ on the interval $[a-r, a + r] $.
- (Integration of power series) For any closed interval $[y, z] $ contained in $(a - R, a +R) $, we have $$\int_{ [y, z] } f(x) \d x = \sum_{ n=0 }^{ \infty } c_n \frac{(z -a)^{n + 1} - (y - a)^{n + 1}}{ n+1 } $$
Consider the power series $\sum_{ n=0 }^{ \infty } n (x -1)^n $. The ratio test shows that this series converges when $\abs{ x - 1 } < 1 $ and diverges when $\abs{ x - 1 } > 1 $. Thus the only possible value for the radius of convergence is $R = 1 $.