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alternating series 

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Decide whether the following series is convergent or not

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If you view the series as 

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\\[\\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\\]

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It is an alternating series. However, alternating series does not work (why not?)

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Instead, observe that 

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\\[\\frac{1}{\\sqrt[\\var{m}]{n}-1} - \\frac{1}{\\sqrt[\\var{m}]{n}+1} = \\frac{1}{n^{2/{\\var{m}}}-1}.\\]

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Also, observe that 

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\\[ \\frac{1}{n^{2/{\\var{m}}}} \\leq \\frac{1}{n^{2/{\\var{m}}}-1}. \\]

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By $p$-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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$\\displaystyle\\sum_{n=2}^\\infty \\left( \\frac{1}{\\sqrt[\\var{m}]{n}-1} - \\frac{1}{\\sqrt[\\var{m}]{n}+1}\\right)$

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