[proofplan] The strategy is to show that the ratio limit $L$ forces the root limit $\limsup_{n \to \infty} |a_n|^{1/n}$ to equal $L$, so that the [Cauchy-Hadamard Formula](/theorems/203) yields $R = 1/L$. The key ingredient is a Cesaro-type lemma: if the consecutive differences of a sequence converge, then the sequence's averages converge to the same limit. We apply this to the logarithms of the coefficients. [/proofplan] [step:Reduce to the finite positive case $0 < L < \infty$] If $L = 0$, then for every $\varepsilon > 0$ we have $|a_{n+1}/a_n| < \varepsilon$ for all $n$ sufficiently large, so $|a_n|^{1/n} \to 0$ by the argument below applied with $\ell = -\infty$ (or by direct comparison with a geometric sequence of ratio $\varepsilon$). The [Cauchy-Hadamard Formula](/theorems/203) gives $R = 1/0 = \infty$. If $L = +\infty$, then $|a_{n+1}/a_n| \to +\infty$, so $|a_n|^{1/n} \to +\infty$ by the same Cesaro-type argument applied to the divergent differences. The Cauchy-Hadamard Formula gives $R = 1/\infty = 0$. For the remainder of the proof, assume $0 < L < \infty$, so that $\log L$ is a finite real number. [/step] [step:Reformulate the ratio hypothesis in terms of logarithms] Since $a_n \neq 0$ for all sufficiently large $n$ and $L > 0$, the quantities $\log |a_n|$ are well-defined for $n \geq N_0$ for some index $N_0 \in \mathbb{N}$. Define the sequence \begin{align*} b_n := \log |a_n|, \quad n \geq N_0. \end{align*} The hypothesis $\lim_{n \to \infty} |a_{n+1}/a_n| = L$ translates, by continuity of the logarithm on $(0, \infty)$, to \begin{align*} \lim_{n \to \infty} (b_{n+1} - b_n) = \log L. \end{align*} The target $\lim_{n \to \infty} |a_n|^{1/n} = L$ is equivalent to $b_n / n \to \log L$, since $|a_n|^{1/n} = e^{b_n/n}$ and the exponential function is continuous. Thus the problem reduces to proving the following purely real-analytic statement. [guided] Why pass to logarithms? The ratio $|a_{n+1}/a_n|$ is a multiplicative comparison of consecutive terms, while the $n$-th root $|a_n|^{1/n}$ is a multiplicative average over the first $n$ terms. Taking logarithms converts both to additive quantities: the ratio becomes a difference $b_{n+1} - b_n$, and the $n$-th root becomes an average $b_n/n$. The additive setting is where Cesaro-type averaging lemmas apply. Since $\lim_{n \to \infty} |a_{n+1}/a_n| = L > 0$, the ratios are eventually bounded away from $0$ and $\infty$, which forces $a_n \neq 0$ for all sufficiently large $n$. Fix such an index $N_0 \in \mathbb{N}$ and define \begin{align*} b_n := \log |a_n|, \quad n \geq N_0. \end{align*} The continuity of $\log : (0, \infty) \to \mathbb{R}$ converts the ratio hypothesis into \begin{align*} \lim_{n \to \infty} (b_{n+1} - b_n) = \lim_{n \to \infty} \log \left| \frac{a_{n+1}}{a_n} \right| = \log L. \end{align*} Similarly, $|a_n|^{1/n} = \exp(b_n / n)$, so $|a_n|^{1/n} \to L$ is equivalent to $b_n / n \to \log L$ by continuity of the exponential. The entire proof now reduces to a single claim about real sequences. [/guided] [/step] [step:Prove the Cesaro-type lemma: convergent differences imply convergent averages] [claim:Cesaro-Type Averaging Lemma] Let $(b_n)_{n \geq N_0}$ be a real sequence. If $\lim_{n \to \infty} (b_{n+1} - b_n) = \ell$ for some $\ell \in \mathbb{R}$, then $\lim_{n \to \infty} b_n / n = \ell$. [/claim] [proof] Let $\varepsilon > 0$. By hypothesis, there exists $N \geq N_0$ such that \begin{align*} |b_{k+1} - b_k - \ell| < \varepsilon \quad \text{for all } k \geq N. \end{align*} Write each difference as $b_{k+1} - b_k = \ell + \alpha_k$ where $|\alpha_k| < \varepsilon$ for all $k \geq N$. Summing from $k = N$ to $k = n - 1$ (for $n > N$) via the telescoping identity yields \begin{align*} b_n - b_N = \sum_{k=N}^{n-1} (b_{k+1} - b_k) = (n - N)\ell + \sum_{k=N}^{n-1} \alpha_k. \end{align*} Rearranging and dividing by $n$: \begin{align*} \frac{b_n}{n} = \frac{b_N}{n} + \frac{n - N}{n} \cdot \ell + \frac{1}{n} \sum_{k=N}^{n-1} \alpha_k. \end{align*} We estimate each term on the right-hand side. The first term $b_N / n \to 0$ as $n \to \infty$ since $b_N$ is a fixed constant. The second term $(n - N)/n \cdot \ell \to \ell$ as $n \to \infty$. For the third term, the triangle inequality and the bound $|\alpha_k| < \varepsilon$ give \begin{align*} \left| \frac{1}{n} \sum_{k=N}^{n-1} \alpha_k \right| \leq \frac{1}{n} \sum_{k=N}^{n-1} |\alpha_k| < \frac{(n - N)}{n} \cdot \varepsilon \leq \varepsilon. \end{align*} Choose $N_1 \geq N$ large enough that $|b_N / n| < \varepsilon$ and $|(n-N)/n \cdot \ell - \ell| < \varepsilon$ for all $n \geq N_1$. Then for all $n \geq N_1$: \begin{align*} \left| \frac{b_n}{n} - \ell \right| &\leq \left| \frac{b_N}{n} \right| + \left| \frac{n-N}{n} \cdot \ell - \ell \right| + \left| \frac{1}{n} \sum_{k=N}^{n-1} \alpha_k \right| < \varepsilon + \varepsilon + \varepsilon = 3\varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $b_n / n \to \ell$. [/proof] [guided] This is a discrete analogue of the fact that if a function's derivative has a [limit](/page/Limit) at infinity, then the function grows linearly at that rate. The idea is that convergent differences "stabilise" the sequence near a linear function $n \mapsto n\ell + C$, and dividing by $n$ extracts the slope. Let $\varepsilon > 0$. The hypothesis $b_{k+1} - b_k \to \ell$ provides an index $N \geq N_0$ such that \begin{align*} |b_{k+1} - b_k - \ell| < \varepsilon \quad \text{for all } k \geq N. \end{align*} Writing $b_{k+1} - b_k = \ell + \alpha_k$ with $|\alpha_k| < \varepsilon$, we telescope from $k = N$ to $k = n-1$: \begin{align*} b_n - b_N = \sum_{k=N}^{n-1}(b_{k+1} - b_k) = (n-N)\ell + \sum_{k=N}^{n-1} \alpha_k. \end{align*} The first term on the right is the "linear approximation" $(n-N)\ell$, while $\sum_{k=N}^{n-1} \alpha_k$ is the accumulated error. Each $|\alpha_k| < \varepsilon$, so the total error grows at most linearly: $|\sum \alpha_k| \le (n-N)\varepsilon$. Dividing by $n$: \begin{align*} \frac{b_n}{n} = \underbrace{\frac{b_N}{n}}_{\to 0} + \underbrace{\frac{n-N}{n}}_{\to 1} \cdot \ell + \frac{1}{n}\sum_{k=N}^{n-1} \alpha_k. \end{align*} The three terms are estimated separately. The first term $b_N/n \to 0$ because $b_N$ is a fixed constant. The second term $(n-N)\ell/n \to \ell$. The error term satisfies \begin{align*} \left|\frac{1}{n}\sum_{k=N}^{n-1}\alpha_k\right| \leq \frac{n-N}{n} \varepsilon \leq \varepsilon. \end{align*} Choosing $N_1 \geq N$ large enough that $|b_N/n| < \varepsilon$ and $|(n-N)/n - 1| \cdot |\ell| < \varepsilon$ for all $n \geq N_1$, the triangle inequality gives \begin{align*} \left|\frac{b_n}{n} - \ell\right| \leq \left|\frac{b_N}{n}\right| + \left|\frac{n-N}{n}\ell - \ell\right| + \left|\frac{1}{n}\sum_{k=N}^{n-1}\alpha_k\right| < 3\varepsilon \end{align*} for all $n \geq N_1$. Since $\varepsilon > 0$ was arbitrary, $b_n/n \to \ell$. [/guided] [/step] [step:Conclude via the Cauchy-Hadamard Formula] Applying the Cesaro-type lemma with $\ell = \log L$ gives $b_n / n \to \log L$. Since the exponential function is continuous, \begin{align*} |a_n|^{1/n} = e^{b_n/n} \to e^{\log L} = L. \end{align*} Because the limit $\lim_{n \to \infty} |a_n|^{1/n} = L$ exists, it coincides with the $\limsup$: \begin{align*} \limsup_{n \to \infty} |a_n|^{1/n} = \lim_{n \to \infty} |a_n|^{1/n} = L. \end{align*} The [Cauchy-Hadamard Formula](/theorems/203) states that $R = 1 / \limsup_{n \to \infty} |a_n|^{1/n}$. Substituting: \begin{align*} R = \frac{1}{L}, \end{align*} with the conventions $1/0 = \infty$ and $1/\infty = 0$. [guided] We have established that $b_n/n \to \log L$, where $b_n = \log |a_n|$ for $n \geq N_0$. To pass back to the $n$-th root, apply the exponential. Since $\exp : \mathbb{R} \to (0,\infty)$ is continuous, the composition $|a_n|^{1/n} = \exp(b_n/n)$ preserves the limit: \begin{align*} \lim_{n \to \infty} |a_n|^{1/n} = \exp\!\left(\lim_{n \to \infty} \frac{b_n}{n}\right) = e^{\log L} = L. \end{align*} When a sequence has a genuine limit, its $\limsup$ and $\liminf$ both equal that limit. In particular, $\limsup_{n \to \infty} |a_n|^{1/n} = L$. The [Cauchy-Hadamard Formula](/theorems/203) asserts that the [radius of convergence](/theorems/265) of $\sum a_n(x-c)^n$ is \begin{align*} R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}. \end{align*} Substituting $\limsup_{n \to \infty} |a_n|^{1/n} = L$ yields $R = 1/L$, as required. [/guided] [/step]