// Numbas version: finer_feedback_settings {"name": "alternating series 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "alternating series 1", "tags": [], "metadata": {"description": "
alternating series
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "Decide whether the following series is convergent or not
", "advice": "If you view the series as
\n\\[\\frac{1}{\\sqrt[\\var{m}]{2}-1} - \\frac{1}{\\sqrt[\\var{m}]{2}+1} + \\frac{1}{\\sqrt[\\var{m}]{3}-1} - \\frac{1}{\\sqrt[\\var{m}]{3}+1} + \\frac{1}{\\sqrt[\\var{m}]{4}-1} - \\frac{1}{\\sqrt[\\var{m}]{4}+1} + \\cdots\\]
\nIt is an alternating series. However, alternating series does not work (why not?)
\n\nInstead, observe that
\n\\[\\frac{1}{\\sqrt[\\var{m}]{n}-1} - \\frac{1}{\\sqrt[\\var{m}]{n}+1} = \\frac{1}{n^{2/{\\var{m}}}-1}.\\]
\nAlso, observe that
\n\\[ \\frac{1}{n^{2/{\\var{m}}}} \\leq \\frac{1}{n^{2/{\\var{m}}}-1}. \\]
\nBy $p$-series test, $\\displaystyle \\sum_{n=2}^\\infty \\frac{1}{n^{2/{\\var{m}}}}$ is divergent (as $2/{\\var{m}} \\leq 1$). Then by comparison test, $\\displaystyle \\sum_{n=2}^\\infty \\frac{1}{n^{2/{\\var{m}}}-1} = \\sum_{n=2}^\\infty \\left(\\frac{1}{\\sqrt[\\var{m}]{n}-1} - \\frac{1}{\\sqrt[\\var{m}]{n}+1}\\right)$ is also divergent.
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