The Stolz-Cesàro theorem provides a useful tool for calculating the limit of a ratio of sequences. It is particularly useful when the denominator tends to infinity and the calculation of the limit is not immediate. This theorem represents a generalization of Cesàro's criteria and is often used to simplify the verification of sequence convergence.
Stolz-Cesàro Theorem. Let \( \{a_n\}_{n \in \mathbb{N}} \) and \( \{b_n\}_{n \in \mathbb{N}} \) be two numerical sequences, with \( \{b_n\}_{n \in \mathbb{N}} \) satisfying the following properties:
- \( b_n > 0 \) for all \( n \in \mathbb{N} \)
- \( b_{n+1} > b_n \) for all \( n \in \mathbb{N} \)
- \(\lim_{n\to \infty } b_n = +\infty \).
If the limit exists:
\[ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L \]
then the limit also exists:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]
Proof. Suppose that the limit:
\[ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L \]
exists and equals \( L \). We must prove that:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]
Consider the definition of limit:
\[ \forall \varepsilon > 0 \,\, \exists n_\varepsilon \in \mathbb{N} : \left| \frac{a_{n+1} - a_n}{b_{n+1} - b_n} - L \right| < \varepsilon \quad \forall n \geq n_\varepsilon \]
Multiply both sides by \( b_{n+1} - b_n \), which is positive:
\[ ( L - \varepsilon)(b_{n+1} - b_n) < a_{n+1} - a_n < ( L + \varepsilon)(b_{n+1} - b_n) \]
Sum this inequality 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 are telescoping, which means that intermediate terms cancel out. In detail:
\[ \sum_{k=n_\varepsilon}^{n-1} (a_{k+1} - a_k) = (a_{n_\varepsilon+1} - a_{n_\varepsilon}) + (a_{n_\varepsilon+2} - a_{n_\varepsilon+1}) + \dots + (a_n - a_{n-1}) = a_n - a_{n_\varepsilon} \]
The same applies to the sum of \( b_k \):
\[ \sum_{k=n_\varepsilon}^{n-1} (b_{k+1} - b_k) = b_n - b_{n_\varepsilon} \]
Therefore, the inequality becomes:
\[ ( L - \varepsilon)(b_n - b_{n_\varepsilon}) < a_n - a_{n_\varepsilon} < ( L + \varepsilon)(b_n - b_{n_\varepsilon}) \]
Divide by \( b_n \)
\[ ( L - \varepsilon) \left( 1 - \frac{b_{n_\varepsilon}}{b_n} \right) < \frac{a_n}{b_n} - \frac{a_{n_\varepsilon}}{b_n} < ( L + \varepsilon) \left( 1 - \frac{b_{n_\varepsilon}}{b_n} \right) \]
and rearrange the terms:
\[ ( L - \varepsilon) + \frac{a_{n_\varepsilon}}{b_n} < \frac{a_n}{b_n} < ( L + \varepsilon) + \frac{a_{n_\varepsilon}}{b_n} \]
Since \( b_n \to +\infty \), it follows that:
\[ \lim_{n \to +\infty} \frac{b_{n_\varepsilon}}{b_n} = 0 \]
Finally, taking the limit of the inequality:
\[ L - \varepsilon \leq \lim_{n \to +\infty} \frac{a_n}{b_n} \leq L + \varepsilon \]
Since \( \varepsilon \) is arbitrary, we conclude that:
\[ \lim_{n \to +\infty} \frac{a_n}{b_n} = L \]
This concludes the proof of the Stolz-Cesàro Theorem.
The proof for the case \( n \to -\infty \) follows the same reasoning used for \( n \to +\infty \), with only one modification: the interval of the telescoping sum.
For \( n \to +\infty \), the sum goes from \( n_\varepsilon \) up to \( n-1 \):
\[ \sum_{k = n_\varepsilon}^{n-1} (a_{k+1} - a_k) = a_n - a_{n_\varepsilon} \]
For \( n \to -\infty \), the sum goes from \( n \) up to \( n_\varepsilon - 1 \):
\[ \sum_{k = n}^{n_\varepsilon - 1} (a_{k+1} - a_k) = a_{n_\varepsilon} - a_n \]
This reflects the fact that \( n \to -\infty \), so \( n \leq n_\varepsilon \).
The remaining part of the proof remains unchanged, leading to the conclusion:
\[ \lim_{n \to -\infty} \frac{a_n}{b_n} = L \]
This concludes the proof for \( n \to -\infty \).
Corollary I. Suppose that the sequence \(\{a_n\}\) is such that:
\[\lim_{n \to \infty} (a_{n+1} - a_n) = L \]
Then:
\[\lim_{n \to \infty} \frac{a_n}{n} = L \]
Proof. This is a special case of the Stolz-Cesàro theorem. Consider the sequence \(\{b_n\}\) defined by \( b_n = n \) and observe that \(b_{n+1} - b_n = 1\), therefore:
\[\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n \to \infty} (a_{n+1} - a_n) = L \]
Finally, by the Stolz-Cesàro theorem:
\[\lim_{n \to \infty} \frac{a_n}{n} = L \]
Corollary II. Let \( \{a_n\}_{n \in \mathbb{N}} \) be a sequence and consider the sequence of its means, defined by
\[ \alpha_n = \frac{1}{n} \sum_{k=0}^{n-1} a_k \quad n \in \mathbb{N} \]
If \( \{a_n\}_{n \in \mathbb{N}} \) converges to the value \( L \), then the sequence \( \{\alpha_n\}_{n \in \mathbb{N}} \) also tends to the same limit:
\[ \lim_{n \to \infty} a_n = L \implies \lim_{n \to \infty} \alpha_n = L \]
Proof. To prove this result, we use the Stolz-Cesàro theorem. Consider the sequences
\[ c_n = \sum_{k=0}^{n-1} a_k \quad b_n = n \quad n \in \mathbb{N} \]
Rewriting the limits, we obtain:
\[ \lim_{n \to \infty} \alpha_n = \lim_{n \to \infty} \frac{c_n}{b_n} \]
Since \( \{b_n\} \) is strictly increasing and unbounded, and \( \{c_n\} \) satisfies the hypotheses of the Stolz-Cesàro theorem, we have:
\[ \lim_{n \to \infty} \frac{c_{n+1} - c_n}{b_{n+1} - b_n} = \lim_{n \to \infty} a_n = L \]
From which it follows that:
\[ \lim_{n \to \infty} \alpha_n = \lim_{n \to \infty} \frac{c_n}{b_n} = L \]
Corollary III. Suppose now that the sequence \( \{a_n\}_{n \in \mathbb{N}} \) has limit \( L \) and that \( a_n > 0 \) for all \( n \). Then the following relation holds:
\[ \lim_{n \to \infty} \sqrt[n]{\prod_{k=0}^{n-1} a_k} = L \]
Proof. Let's define a new sequence
\[ b_n = \log a_n \quad n \in \mathbb{N} \]
The sequence \( \{b_n\} \) converges to \( \log L \). The arithmetic means of \( \{b_n\} \) are
\[ \beta_n = \frac{1}{n} \sum_{k=0}^{n-1} b_k = \frac{1}{n} \sum_{k=0}^{n-1} \log a_k = \log \sqrt[n]{\prod_{k=0}^{n-1} a_k}. \]
Applying Corollary II, we know that
\[ \lim_{n \to \infty} \beta_n = \lim_{n \to \infty} b_n = \log L, \]
which implies
\[ \lim_{n \to \infty} \sqrt[n]{\prod_{k=0}^{n-1} a_k} = L \]
Corollary IV. Let \(\{a_n\}_{n \in \mathbb{N}}\) be a sequence of strictly positive real numbers. The following result holds:
If \(\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L\) exists, then \(\displaystyle \lim_{n \to \infty} \sqrt[n]{a_n} = L\) also exists.
Proof. Consider a new auxiliary sequence \(\{b_n\}_{n \in \mathbb{N}}\) defined in the following way:
\[b_n = \frac{a_n}{a_{n-1}} \quad \text{for } n \geq 1 \quad \text{and} \quad b_0 = a_0\]
By Corollary III, we know that:
\[\lim_{n \to \infty} \sqrt[n]{\prod_{k=0}^n b_k} = \lim_{n \to \infty} b_n = L\]
Now observe that the telescopic product of \(b_k\) can be rewritten as:
\[\sqrt[n]{\prod_{k=0}^n b_k} = \sqrt[n]{\prod_{k=1}^n \frac{a_k}{a_{k-1}} \cdot a_0} = \sqrt[n]{a_n}\]
From which the thesis immediately follows.
