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  "body": "\n## Proof\n\nWe compare each term $\\frac{1}{k}$ with the integral of $\\frac{1}{t}$ over $[k, k + 1]$, and use the monotone convergence theorem.\n\n1. For each integer $n \\ge 1$, let $H_n$ be the $n$-th [[harmonic number|0D2Y6G4C4S]], and let $a_n = H_n - \\ln n$. Here $\\ln$ is the [[natural logarithm|0044CPHJBV]].\n2. Let $k \\ge 1$ be an integer. By the [[integral of 1/x|0K1N1Y4EGZ]] and the [[logarithm of a quotient|0J7PHRBNNS]], $\\int_k^{k+1} \\frac{dt}{t} = \\ln \\frac{k+1}{k} = \\ln(k + 1) - \\ln k$.\n3. The function $t \\mapsto \\frac{1}{t}$ is continuous on $[k, k + 1]$, by [[continuity of quotients|0GR2APWTAC]]. So it is Riemann integrable there, since [[continuous functions are Riemann integrable|06SESV7GQY]].\n4. For a constant $c$, the [[integral of a power|0Q3HYFT9RQ]] with exponent $0$ and [[linearity of the integral|080RWVVSB6]] give $\\int_k^{k+1} c \\, dt = c$.\n5. For $t \\in [k, k + 1]$, we have $\\frac{1}{k + 1} \\le \\frac{1}{t} \\le \\frac{1}{k}$.\n6. By [[monotonicity of the integral|0MBWQKM8QR]] and steps 2 to 5,\n$$\\frac{1}{k + 1} \\le \\ln(k + 1) - \\ln k \\le \\frac{1}{k}.$$\n7. For $n \\ge 1$, $a_{n+1} - a_n = \\frac{1}{n + 1} - (\\ln(n + 1) - \\ln n)$. This is $\\le 0$ by the left inequality of step 6 with $k = n$. So $(a_n)$ is [[decreasing|0KN22AAW5R]].\n8. Since $e^0 = 1$, the definition of $\\ln$ gives $\\ln 1 = 0$.\n9. Summing the right inequality of step 6 over $k = 1, \\dots, n$ gives a telescoping sum: $\\ln(n + 1) - \\ln 1 \\le H_n$. By step 8, $\\ln(n + 1) \\le H_n$.\n10. By step 9 and step 6 with $k = n$, $a_n \\ge \\ln(n + 1) - \\ln n \\ge \\frac{1}{n + 1} \u003e 0$. So $(a_n)$ is bounded below by $0$.\n11. The sequence $(-a_n)$ is increasing by step 7, and bounded above by $0$ by step 10.\n12. By the [[monotone convergence theorem|0FM6N4R0VB]], $(-a_n)$ converges. So $(a_n)$ converges to some real number $\\gamma$.\n13. By step 8, $a_1 = 1 - \\ln 1 = 1$. By steps 7 and 10, $0 \\le a_n \\le 1$ for all $n \\ge 1$.\n14. Since [[limits preserve non-strict inequalities|0N32V0E3XJ]], step 13 gives $0 \\le \\gamma \\le 1$. So the limit exists and lies in $[0, 1]$.\n15. It remains to show $H_n \\sim \\ln n$. Let $\\varepsilon \u003e 0$. By the [[Archimedean property|0A74SX55QP]], there is an integer $N \\ge 2$ with $N \u003e e^{1/\\varepsilon}$.\n16. Let $n \\ge N$. Suppose that $\\ln n \u003c \\frac{1}{\\varepsilon}$. Since the [[exponential function is strictly increasing|0KN2V7840E]], $n = e^{\\ln n} \u003c e^{1/\\varepsilon} \u003c N$. This contradicts $n \\ge N$.\n17. So $\\ln n \\ge \\frac{1}{\\varepsilon} \u003e 0$ for all $n \\ge N$. With step 13, $|H_n - \\ln n| = a_n \\le 1 \\le \\varepsilon \\ln n$ for all $n \\ge N$.\n18. By the definition of [[little o notation|04TSHB9H8D]], step 17 says $H_n - \\ln n = o(\\ln n)$. This is the definition of [[asymptotic equivalence|00SB5EA3VW]], so $H_n \\sim \\ln n$.\n\n## Idea\n\nThe integral $\\int_k^{k+1} \\frac{dt}{t}$ lies between $\\frac{1}{k+1}$ and $\\frac{1}{k}$. So the gap $H_n - \\ln n$ loses a little at each step, but it never drops below $0$. A decreasing sequence that is bounded below converges. Since the gap stays bounded while $\\ln n \\to \\infty$, the two sequences are asymptotically equal.\n",
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