Proof of the Stolz–Cesàro theorem

Proof

We sum the hypothesis over a block of indices, so that the differences telescope.

1. Let $\varepsilon > 0$. By the definition of the limit of a sequence, there is a natural number $N_1$ such that $$\left| \frac{a_{k+1} - a_k}{b_{k+1} - b_k} - L \right| < \varepsilon \quad \text{for all } k \ge N_1.$$
2. Since $b_n \to +\infty$ in the sense of a sequence tending to infinity, there is $N_2$ with $b_n > 0$ for all $n \ge N_2$. Let $N = \max(N_1, N_2)$, so $b_N > 0$.
3. Since $(b_n)$ is strictly increasing, as a monotone sequence, we have $b_{k+1} - b_k > 0$ for every $k$.
4. Multiply the inequality of step 1 by $b_{k+1} - b_k > 0$. For each $k \ge N$ this gives $$(L - \varepsilon)(b_{k+1} - b_k) < a_{k+1} - a_k < (L + \varepsilon)(b_{k+1} - b_k).$$
5. Let $n > N$. Add the inequalities of step 4 for $k = N, N+1, \dots, n-1$. Both sides telescope, and we get $$(L - \varepsilon)(b_n - b_N) < a_n - a_N < (L + \varepsilon)(b_n - b_N).$$
6. Subtract $L(b_n - b_N)$ from each part of step 5. This gives $\left| a_n - a_N - L(b_n - b_N) \right| < \varepsilon (b_n - b_N)$.
7. For $n > N$ we have $b_n > b_N > 0$, so $b_n \ne 0$ and $0 < b_n - b_N < b_n$. We can write $$\frac{a_n}{b_n} - L = \frac{a_N - L b_N}{b_n} + \frac{a_n - a_N - L(b_n - b_N)}{b_n}.$$
8. By steps 6 and 7, the second fraction in step 7 has absolute value less than $\varepsilon (b_n - b_N)/b_n < \varepsilon$.
9. Let $c = \abs{a_N - L b_N}$; it does not depend on $n$. Since $b_n \to +\infty$, there is $N_3$ with $b_n > c/\varepsilon$ for all $n \ge N_3$. For such $n$ with $b_n > 0$, the first fraction has absolute value $c / b_n < \varepsilon$.
10. Let $n \ge \max(N + 1, N_3)$. By the triangle inequality and steps 8 and 9, $\abs{a_n / b_n - L} < 2\varepsilon$.
11. Since $\varepsilon > 0$ was arbitrary, $a_n / b_n \to L$ as $n \to \infty$.