The Stolz–Cesàro theorem provides a fundamental tool for computing the limit of ratios of sequences. It is especially useful when the denominator tends to infinity and a direct evaluation of the limit proves awkward or leads to an indeterminate form.
This result may be regarded as a generalisation of Cesàro's theorem on arithmetic means, and it is widely employed in the study of the convergence of sequences.
Throughout, we adopt the convention that \( \mathbb{N} = \{0,1,2,\dots\} \).
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Stolz–Cesàro Theorem. Let \( \{a_n\}_{n \in \mathbb{N}} \) and \( \{b_n\}_{n \in \mathbb{N}} \) be two sequences of real numbers. Suppose that:
- \( b_n > 0 \) for all sufficiently large \( n \);
- \( b_{n+1} > b_n \) for all sufficiently large \( n \);
- \[ \lim_{n \to \infty} b_n = +\infty. \]
If the limit
\[ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L \in \mathbb{R} \]
exists, then so does the limit
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L. \]
There are also variants of the Stolz–Cesàro theorem for the case in which the limit of the ratio of the increments equals \(+\infty\) or \(-\infty\), as well as versions suited to certain indeterminate forms of the type \(\displaystyle \frac{0}{0}\). In this text we concentrate on the most commonly used form, namely the one pertaining to the indeterminate form \(\displaystyle \frac{\infty}{\infty}\) with a finite real limit.
Proof. Suppose that
\[ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L. \]
We wish to prove that
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L. \]
By the definition of limit, for every \( \varepsilon > 0 \) there exists \( n_\varepsilon \in \mathbb{N} \), chosen large enough to guarantee also that \( b_n > 0 \) and \( b_{n+1} > b_n \) for every \( n \ge n_\varepsilon \), such that
\[ \left| \frac{a_{n+1} - a_n}{b_{n+1} - b_n} - L \right| < \varepsilon, \qquad \forall n \ge n_\varepsilon. \]
Equivalently,
\[ L - \varepsilon < \frac{a_{n+1} - a_n}{b_{n+1} - b_n} < L + \varepsilon. \]
Since \( b_{n+1} - b_n > 0 \) for every \( n \ge n_\varepsilon \), we may multiply through the inequality to obtain:
\[ (L - \varepsilon) (b_{n+1} - b_n) < a_{n+1} - a_n < (L + \varepsilon) (b_{n+1} - b_n). \]
Summing term by term from \( k = n_\varepsilon \) to \( k = n - 1 \):
\[ \sum_{k=n_\varepsilon}^{n-1} (L - \varepsilon)(b_{k+1} - b_k) < \sum_{k=n_\varepsilon}^{n-1} (a_{k+1} - a_k) < \sum_{k=n_\varepsilon}^{n-1} (L + \varepsilon)(b_{k+1} - b_k). \]
The sums telescope. Indeed:
\[ \sum_{k=n_\varepsilon}^{n-1} (a_{k+1} - a_k) = a_n - a_{n_\varepsilon}, \]
and, similarly,
\[ \sum_{k=n_\varepsilon}^{n-1} (b_{k+1} - b_k) = b_n - b_{n_\varepsilon}. \]
Therefore:
\[ (L - \varepsilon) (b_n - b_{n_\varepsilon}) < a_n - a_{n_\varepsilon} < (L + \varepsilon) (b_n - b_{n_\varepsilon}). \]
Dividing by \( b_n > 0 \), we obtain:
\[ (L - \varepsilon) \left( 1 - \frac{b_{n_\varepsilon}}{b_n} \right) + \frac{a_{n_\varepsilon}}{b_n} < \frac{a_n}{b_n} < (L + \varepsilon) \left( 1 - \frac{b_{n_\varepsilon}}{b_n} \right) + \frac{a_{n_\varepsilon}}{b_n}. \]
Since
\[ \lim_{n \to \infty} \frac{b_{n_\varepsilon}}{b_n} = 0, \qquad \lim_{n \to \infty} \frac{a_{n_\varepsilon}}{b_n} = 0, \]
passing to the lower and upper limits in the preceding inequality we obtain:
\[ L - \varepsilon \le \liminf_{n \to \infty} \frac{a_n}{b_n} \le \limsup_{n \to \infty} \frac{a_n}{b_n} \le L + \varepsilon. \]
Since \( \varepsilon > 0 \) is arbitrary, it follows that
\[ \liminf_{n \to \infty} \frac{a_n}{b_n} = \limsup_{n \to \infty} \frac{a_n}{b_n} = L. \]
Hence:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L. \]
This completes the proof of the Stolz–Cesàro theorem.
Corollary I. If
\[ \lim_{n \to \infty} (a_{n+1} - a_n) = L, \]
then
\[ \lim_{n \to \infty} \frac{a_n}{n} = L. \]
Proof. It suffices to apply the Stolz–Cesàro theorem to the sequence \( b_n = n \). The ratio \(a_n/n\) is, of course, considered for \(n\ge 1\). Indeed:
\[ b_{n+1} - b_n = 1, \]
hence
\[ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} (a_{n+1} - a_n) = L. \]
Therefore:
\[ \lim_{n \to \infty} \frac{a_n}{n} = L. \]
Corollary II (Cesàro's Theorem). Let \( \{a_n\}_{n \in \mathbb{N}} \) be a sequence converging to \( L \). For every \(n\ge 1\), define:
\[ \alpha_n = \frac{1}{n} \sum_{k=0}^{n-1} a_k. \]
Then:
\[ \lim_{n \to \infty} \alpha_n = L. \]
Proof. Put
\[ c_n = \sum_{k=0}^{n-1} a_k, \qquad b_n = n. \]
Then:
\[ \alpha_n = \frac{c_n}{b_n}. \]
Moreover:
\[ \frac{c_{n+1} - c_n}{b_{n+1} - b_n} = \frac{a_n}{1} = a_n. \]
Since \( a_n \to L \), the Stolz–Cesàro theorem yields:
\[ \lim_{n \to \infty} \alpha_n = \lim_{n \to \infty} \frac{c_n}{b_n} = L. \]
Corollary III. Let \( \{a_n\}_{n \in \mathbb{N}} \) be a sequence such that:
- \( a_n > 0 \) for every \( n \);
- \[ \lim_{n \to \infty} a_n = L > 0. \]
Then, for \(n\ge 1\):
\[ \lim_{n \to \infty} \sqrt[n]{\prod_{k=0}^{n-1} a_k} = L. \]
Proof. Define:
\[ u_n = \log a_n. \]
Since \( a_n \to L > 0 \), we have:
\[ u_n = \log a_n \longrightarrow \log L. \]
Consider the arithmetic means:
\[ \beta_n = \frac{1}{n} \sum_{k=0}^{n-1} u_k, \qquad n\ge 1. \]
Substituting the definition of \( u_k \), we obtain:
\[ \beta_n = \frac{1}{n} \sum_{k=0}^{n-1} \log a_k = \log \left( \sqrt[n]{\prod_{k=0}^{n-1} a_k} \right). \]
By Corollary II:
\[ \lim_{n \to \infty} \beta_n = \log L. \]
Applying the exponential function:
\[ \lim_{n \to \infty} \sqrt[n]{\prod_{k=0}^{n-1} a_k} = L. \]
Corollary IV. Let \( \{a_n\}_{n \in \mathbb{N}} \) be a sequence of strictly positive real numbers.
If
\[ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L > 0, \]
then, for \(n\ge 1\):
\[ \lim_{n \to \infty} \sqrt[n]{a_n} = L. \]
Proof. For every \( n \ge 1 \), define
\[ b_n = \frac{a_n}{a_{n-1}}. \]
By hypothesis:
\[ b_n \to L. \]
Applying Corollary III to the sequence \( \{b_n\}_{n \ge 1} \), or equivalently to this sequence reindexed from \(0\), we obtain:
\[ \lim_{n \to \infty} \sqrt[n]{\prod_{k=1}^{n} b_k} = L. \]
On the other hand,
\[ \prod_{k=1}^{n} b_k = \prod_{k=1}^{n} \frac{a_k}{a_{k-1}} = \frac{a_n}{a_0}. \]
Hence:
\[ \sqrt[n]{\prod_{k=1}^{n} b_k} = \sqrt[n]{\frac{a_n}{a_0}} = \frac{\sqrt[n]{a_n}}{\sqrt[n]{a_0}}. \]
Since
\[ \lim_{n \to \infty} \sqrt[n]{a_0} = 1, \]
it follows that:
\[ \lim_{n \to \infty} \sqrt[n]{a_n} = L. \]
This completes the proof.