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  "body": "\n## Proof\n\nFor $n \\ge 1$, write $n^{-p} = e^{-p \\ln n}$, as in the definition of [[real exponentiation|0TV07AQT62]].\n\n1. The [[natural logarithm|0044CPHJBV]] is strictly increasing. Indeed, if $0 \u003c x \u003c y$ and $\\ln x \\ge \\ln y$, then $x = e^{\\ln x} \\ge e^{\\ln y} = y$, since [[exp is strictly increasing|0KN2V7840E]]. This is impossible.\n2. Since $e^0 = 1$, we have $\\ln 1 = 0$. By step 1, $\\ln n \\ge 0$ for $n \\ge 1$, and $\\ln 2 \u003e 0$.\n3. Case $p \\le 0$. For $n \\ge 1$, $-p \\ln n \\ge 0$, so $n^{-p} \\ge e^0 = 1$ by the monotonicity of exp. The terms do not tend to $0$, so the series diverges by the [[term test|0VVY4H92G2]].\n4. Case $p \u003e 0$. If $1 \\le m \u003c n$, then $-p \\ln m \u003e -p \\ln n$ by step 1. So $m^{-p} \u003e n^{-p}$, and the terms $a_n = n^{-p}$ form a decreasing sequence of positive numbers.\n5. By the [[Cauchy condensation test|0B96KMDDD1]], $\\sum a_n$ converges if and only if $\\sum_{k \\ge 0} 2^k a_{2^k}$ converges.\n6. By the [[logarithm of a product|0K2FNY6DYV]], $\\ln(2^k) = k \\ln 2$. So $a_{2^k} = e^{-pk \\ln 2}$ and $2^k = e^{k \\ln 2}$.\n7. By the [[exponential of a sum|0E8HNVBCSP]], $2^k a_{2^k} = e^{k(1-p)\\ln 2} = r^k$ with $r = e^{(1-p)\\ln 2} \u003e 0$.\n8. If $p \u003e 1$, then $(1-p)\\ln 2 \u003c 0$ by step 2, so $0 \u003c r \u003c e^0 = 1$. The [[geometric series|01X5CVV1Q5]] $\\sum r^k$ converges. By step 5, $\\sum n^{-p}$ converges.\n9. If $0 \u003c p \\le 1$, then $(1-p) \\ln 2 \\ge 0$, so $r \\ge 1$ and $r^k \\ge 1$ for all $k$. The condensed series diverges by the term test. By step 5, $\\sum n^{-p}$ diverges.\n10. Steps 3, 8 and 9 show that the series [[converges|0CVVKY7GY2]] exactly when $p \u003e 1$.\n\n## Idea\n\nFor $p \\le 0$ the terms do not even tend to $0$. For $p \u003e 0$ the terms decrease, so Cauchy condensation applies. It turns the series into the geometric series with ratio $2^{1-p}$, which converges exactly when this ratio is below $1$, that is, when $p \u003e 1$.\n",
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