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  "body": "\n## Proof\n\nWe sum the hypothesis over a block of indices, so that the differences telescope.\n\n1. Let $\\varepsilon \u003e 0$. By the definition of the [[limit of a sequence|0XKWSSY8P2]], there is a natural number $N_1$ such that\n$$\\left| \\frac{a_{k+1} - a_k}{b_{k+1} - b_k} - L \\right| \u003c \\varepsilon \\quad \\text{for all } k \\ge N_1.$$\n2. Since $b_n \\to +\\infty$ in the sense of a [[sequence tending to infinity|09AZCV6GMC]], there is $N_2$ with $b_n \u003e 0$ for all $n \\ge N_2$. Let $N = \\max(N_1, N_2)$, so $b_N \u003e 0$.\n3. Since $(b_n)$ is strictly increasing, as a [[monotone sequence|0KN22AAW5R]], we have $b_{k+1} - b_k \u003e 0$ for every $k$.\n4. Multiply the inequality of step 1 by $b_{k+1} - b_k \u003e 0$. For each $k \\ge N$ this gives\n$$(L - \\varepsilon)(b_{k+1} - b_k) \u003c a_{k+1} - a_k \u003c (L + \\varepsilon)(b_{k+1} - b_k).$$\n5. Let $n \u003e N$. Add the inequalities of step 4 for $k = N, N+1, \\dots, n-1$. Both sides telescope, and we get\n$$(L - \\varepsilon)(b_n - b_N) \u003c a_n - a_N \u003c (L + \\varepsilon)(b_n - b_N).$$\n6. Subtract $L(b_n - b_N)$ from each part of step 5. This gives $\\left| a_n - a_N - L(b_n - b_N) \\right| \u003c \\varepsilon (b_n - b_N)$.\n7. For $n \u003e N$ we have $b_n \u003e b_N \u003e 0$, so $b_n \\ne 0$ and $0 \u003c b_n - b_N \u003c b_n$. We can write\n$$\\frac{a_n}{b_n} - L = \\frac{a_N - L b_N}{b_n} + \\frac{a_n - a_N - L(b_n - b_N)}{b_n}.$$\n8. By steps 6 and 7, the second fraction in step 7 has absolute value less than $\\varepsilon (b_n - b_N)/b_n \u003c \\varepsilon$.\n9. Let $c = \\abs{a_N - L b_N}$; it does not depend on $n$. Since $b_n \\to +\\infty$, there is $N_3$ with $b_n \u003e c/\\varepsilon$ for all $n \\ge N_3$. For such $n$ with $b_n \u003e 0$, the first fraction has absolute value $c / b_n \u003c \\varepsilon$.\n10. Let $n \\ge \\max(N + 1, N_3)$. By the triangle inequality and steps 8 and 9, $\\abs{a_n / b_n - L} \u003c 2\\varepsilon$.\n11. Since $\\varepsilon \u003e 0$ was arbitrary, $a_n / b_n \\to L$ as $n \\to \\infty$.\n\n## Idea\n\nThe hypothesis says that each step $a_{k+1} - a_k$ is close to $L$ times the step $b_{k+1} - b_k$. Adding many steps shows that $a_n$ grows like $L b_n$, up to an error of at most $\\varepsilon b_n$ and a fixed constant. Dividing by $b_n$, which tends to infinity, makes the constant negligible.\n",
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