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  "body": "\n## Proof\n\n1. Since $f$ is monotone, it is [[Riemann integrable|04YKAYDQPW]] on $[N, b]$ for each $b \u003e N$. For $b \\ge N$ let $F(b) = \\int_N^b f(x)\\, dx$, with $F(N) = 0$.\n2. $F$ is increasing. For $N \\le b \u003c b'$, [[additivity over intervals|05RCDKHZM6]] gives $F(b') - F(b) = \\int_b^{b'} f(x)\\, dx$. This is at least $0$ by [[monotonicity of the integral|0MBWQKM8QR]], since $f \\ge 0$.\n3. Let $s_M = \\sum_{n=N}^{M} f(n)$ for integers $M \\ge N$. Since $f \\ge 0$, by the [[criterion for nonnegative series|0MX1HZBBXJ]], the series converges if and only if $(s_M)$ is bounded above.\n4. Let $M \u003e N$ and apply the [[comparison of a series and an integral|069HKV4X1E]] with $a = N$ and $n = M - N$. It gives $s_M - f(N) \\le F(M) \\le s_{M-1}$.\n5. Claim: the [[improper integral|0VCTBJMEJF]] converges if and only if $F$ is bounded above on $[N, \\infty)$.\n6. Assume first that the improper integral converges, with value $I$. By the definition of the [[limit at infinity|0Y12QYTJK3]] with $\\varepsilon = 1$, there is $b_1 \\ge N$ with $F(b) \u003c I + 1$ for all $b \u003e b_1$.\n7. For $N \\le b \\le b_1$, step 2 gives $F(b) \\le F(b_1 + 1) \u003c I + 1$. So $F$ is bounded above by $I + 1$.\n8. Assume now that $F$ is bounded above, and let $I = \\sup \\set{F(b) : b \\ge N}$. Let $\\varepsilon \u003e 0$.\n9. By the [[approximation property of the supremum|0M8TGGP9JP]], there is $b_0 \\ge N$ with $I - \\varepsilon \u003c F(b_0)$. For $b \u003e b_0$, step 2 gives $I - \\varepsilon \u003c F(b_0) \\le F(b) \\le I$.\n10. So $\\abs{F(b) - I} \u003c \\varepsilon$ for all $b \u003e b_0$. Thus $\\lim_{b \\to \\infty} F(b) = I$, and the claim of step 5 is proved.\n11. Assume the series converges, and let $S$ be its sum, so $s_M \\le S$ for all $M$. Let $b \\ge N$ and let $M$ be an integer with $M \u003e b$ and $M \u003e N$. By steps 2 and 4, $F(b) \\le F(M) \\le s_{M-1} \\le S$.\n12. So $F$ is bounded above, and the improper integral converges by step 5.\n13. Conversely, assume the improper integral converges. By step 5, $F(b) \\le B$ for all $b \\ge N$, for some real number $B$.\n14. By step 4, $s_M \\le f(N) + F(M) \\le f(N) + B$ for all $M \u003e N$, and $s_N = f(N)$. So $(s_M)$ is bounded above, and the series converges by step 3.\n\n## Idea\n\nThe integral of $f$ from $N$ to $M$ lies between two partial sums of the series, which differ only by one term. Both the partial sums and the integrals increase, since $f \\ge 0$. So one of them stays bounded exactly when the other does, and bounded increasing quantities converge.\n",
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