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  "body": "\n## Statement\n\nLet $p$ be a real number.\nThe series $\\sum_{n=1}^{\\infty} \\frac{1}{n^p}$ converges if and only if $p \u003e 1$.\n\n## Remarks\n\nFor $p = 1$ it is the [[harmonic series|0CXTMAY5PX]], which diverges. For $p \\le 0$ the terms do not tend to $0$.\nOne proof uses the [[Cauchy condensation test|0B96KMDDD1]]. The condensed series is $\\sum 2^n \\cdot 2^{-np} = \\sum (2^{1-p})^n$, a geometric series with ratio $2^{1-p}$.\nAnother proof uses the [[integral test|019QZ6RRVH]] with $f(x) = x^{-p}$.\nFor $p \u003e 1$ the sum is the value $\\zeta(p)$ of the Riemann zeta function. For example, $\\zeta(2) = \\pi^2/6$, which is the [[Basel problem|0DMY75D6FV]].\n\n## Explanations\n\n### L1 human\n\nThe terms $1/n^p$ shrink faster when $p$ is larger.\nThe value $p = 1$ is the border. Above it the sum is finite, and at it or below it the sum is infinite.\n",
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