[proofplan]
We set $\alpha = \limsup_{n \to \infty} |a_n|^{1/n}$ and $R^{*} = 1/\alpha$, then show that the series converges absolutely when $|x - c| < R^{*}$ and diverges when $|x - c| > R^{*}$. The convergence direction uses the defining property of the $\limsup$ to dominate the general term by a geometric series; the divergence direction uses the same property to show the terms fail to tend to zero. By the characterisation of the [radius of convergence](/theorems/202), the two regions force $R = R^{*}$.
[/proofplan]
[step:Set up the candidate radius $R^{*} = 1/\alpha$]
Define $\alpha = \limsup_{n \to \infty} |a_n|^{1/n} \in [0, \infty]$ and set $R^{*} = 1/\alpha$, where $1/0 = \infty$ and $1/\infty = 0$.
[/step]
[step:Show absolute convergence when $|x - c| < R^{*}$ by domination with a geometric series]
Suppose $|x - c| < R^{*}$, so that $|x - c| \cdot \alpha < 1$. Choose $r \in \mathbb{R}$ satisfying $|x - c| \cdot \alpha < r < 1$. Since $\limsup_{n \to \infty} |a_n|^{1/n} = \alpha < r / |x - c|$, the definition of the $\limsup$ guarantees an index $N \in \mathbb{N}$ such that
\begin{align*}
|a_n|^{1/n} < \frac{r}{|x - c|}
\end{align*}
for every $n \geq N$. Raising both sides to the $n$-th power and multiplying by $|x - c|^n$ gives
\begin{align*}
|a_n(x - c)^n| = \bigl(|a_n|^{1/n} |x - c|\bigr)^n < r^n
\end{align*}
for every $n \geq N$. Since $0 < r < 1$, the geometric series $\sum_{n=0}^{\infty} r^n$ converges. By the [Comparison Test](/theorems/173), applied with $|a_n(x-c)^n| \leq r^n$ for $n \geq N$, the series $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges absolutely.
[guided]
We want to show that the [power series](/page/Power%20Series) converges absolutely whenever $|x - c|$ is strictly less than $R^{*}$. The strategy is to squeeze the general term between $0$ and the terms of a convergent geometric series, then invoke the [Comparison Test](/theorems/173).
Suppose $|x - c| < R^{*} = 1/\alpha$. Rearranging, $|x - c| \cdot \alpha < 1$. The strict inequality $|x - c| \cdot \alpha < 1$ leaves room to insert a parameter: choose $r \in \mathbb{R}$ with $|x - c| \cdot \alpha < r < 1$. Why introduce $r$? Because the $\limsup$ characterises the eventual behaviour of the sequence $|a_n|^{1/n}$: it is the infimum of the eventual suprema, so $|a_n|^{1/n}$ eventually lies below any value strictly greater than $\alpha$. Here $r / |x - c| > \alpha$, so there exists $N \in \mathbb{N}$ such that
\begin{align*}
|a_n|^{1/n} < \frac{r}{|x - c|}
\end{align*}
for every $n \geq N$. This is the defining property of $\limsup$: all but finitely many terms of the sequence lie below any value exceeding the $\limsup$.
Now raise both sides to the $n$-th power. Since $|a_n|^{1/n} \geq 0$ and $r/|x - c| > 0$, the inequality is preserved:
\begin{align*}
|a_n| < \left(\frac{r}{|x - c|}\right)^n = \frac{r^n}{|x - c|^n}.
\end{align*}
Multiplying both sides by $|x - c|^n$:
\begin{align*}
|a_n(x - c)^n| = |a_n| \cdot |x - c|^n < r^n.
\end{align*}
Since $0 < r < 1$, the geometric series $\sum_{n=0}^{\infty} r^n = 1/(1 - r) < \infty$ converges. The [Comparison Test](/theorems/173) requires non-negative terms $0 \leq |a_n(x-c)^n| \leq r^n$ for all $n \geq N$, which we have just established. We conclude that $\sum_{n=N}^{\infty} |a_n(x - c)^n|$ converges, and therefore $\sum_{n=0}^{\infty} a_n(x - c)^n$ converges absolutely (a finite number of initial terms does not affect convergence).
[/guided]
[/step]
[step:Show divergence when $|x - c| > R^{*}$ via the failure of the necessary condition for convergence]
Suppose $|x - c| > R^{*}$, so that $|x - c| \cdot \alpha > 1$. Since $\limsup_{n \to \infty} |a_n|^{1/n} = \alpha > 1/|x - c|$, the definition of the $\limsup$ guarantees that $|a_n|^{1/n} > 1/|x - c|$ for infinitely many $n$. For each such $n$,
\begin{align*}
|a_n(x - c)^n| = \bigl(|a_n|^{1/n} |x - c|\bigr)^n > 1^n = 1.
\end{align*}
Therefore $a_n(x - c)^n \not\to 0$ as $n \to \infty$, and the series $\sum_{n=0}^{\infty} a_n(x - c)^n$ diverges.
[guided]
We now show the series diverges whenever $|x - c|$ exceeds $R^{*}$. The argument does not require a comparison with another series — instead, we show that the terms themselves fail to tend to zero, which is a necessary condition for convergence of any series.
Suppose $|x - c| > R^{*} = 1/\alpha$. Then $\alpha > 1/|x - c|$. By the characterisation of $\limsup$ as the largest accumulation point of the sequence $(|a_n|^{1/n})_{n=0}^{\infty}$, there exist infinitely many indices $n$ for which $|a_n|^{1/n}$ exceeds $1/|x - c|$. (In contrast to the convergence direction, where we used the fact that all but finitely many terms lie *below* any value exceeding the $\limsup$, we now use the dual fact: infinitely many terms lie *above* any value strictly below the $\limsup$.)
For each such index $n$, $|a_n|^{1/n} \cdot |x - c| > 1$. Raising to the $n$-th power:
\begin{align*}
|a_n(x - c)^n| = \bigl(|a_n|^{1/n} |x - c|\bigr)^n > 1.
\end{align*}
Since the inequality $|a_n(x - c)^n| > 1$ holds for infinitely many $n$, the general term does not converge to zero. A convergent series must have general term tending to zero, so $\sum_{n=0}^{\infty} a_n(x - c)^n$ diverges.
[/guided]
[/step]
[step:Conclude $R = R^{*}$ from the characterisation of the radius of convergence]
The preceding two steps establish: the series converges absolutely for every $x$ with $|x - c| < R^{*}$, and diverges for every $x$ with $|x - c| > R^{*}$. By the [Radius of Convergence Theorem](/theorems/202), the [radius of convergence](/theorems/265) $R$ is the unique element of $[0, \infty]$ separating the region of absolute convergence from the region of divergence. Since $R^{*}$ satisfies this characterisation, $R = R^{*} = 1/\alpha$.
[/step]
