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alternating series 

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Decide whether the following series is convergent or not

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It may not seem immediate, but we can use alternating series test. It is more visable when we write the first few terms of the series. 

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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) =
\\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} + \\ldots\\]

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It is clear that this is an alternating series where we are summing the terms of the sequence 

\n

\\[u_n =\\left\\lbrace \\begin{array}{rl}
\\frac{1}{\\sqrt[\\var{m}]{n}-1},  &\\mbox{ if $n$  is even}\\\\ 
-\\frac{1}{\\sqrt[\\var{m}]{n-1}+1}, &\\mbox{ if $n$ is odd}
\\end{array}
\\right .\\]

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with $n\\geq 2$. 

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We need to check that $|u_{n+1}|\\leq |u_n|$. We have two cases:

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    \n
  1. If $n$ is even. Then $n+1$ is odd and we have
    \\[|u_{n}| = \\frac{1}{\\sqrt[\\var{m}]{n}-1} \\mbox{ and } |u_{n+1}| = \\frac{1}{\\sqrt[\\var{m}]{n}+1}.\\] Clearly $\\frac{1}{\\sqrt[\\var{m}]{n}+1} \\leq \\frac{1}{\\sqrt[\\var{m}]{n}-1}$. I.e. $|u_{n+1}| \\leq u_n$.
    2. \n
  2. If $n$ is odd. Then $n+1$ is even and we have \\[|u_{n}| = \\frac{1}{\\sqrt[\\var{m}]{n-1}-1} \\mbox{ and } |u_{n+1}| = \\frac{1}{\\sqrt[\\var{m}]{n+1}+1}.\\] Clearly $\\frac{1}{\\sqrt[\\var{m}]{n+1}+1} \\leq \\frac{1}{\\sqrt[\\var{m}]{n-1}-1}$. I.e. $|u_{n+1}| \\leq u_n$.
    3. \n
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Finally, we need to check that $\\displaystyle\\lim_{n\\to\\infty}u_n = 0$. Let $\\varepsilon>0$. Take $N > \\left(\\frac{1}{\\varepsilon} + 1\\right)^{\\var{m}}  $. Then for all $n>N$ we get 

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    \n
  1. $\\left| \\frac{1}{\\sqrt[\\var{m}]{n}-1}\\right| < \\frac{1}{\\sqrt[\\var{m}]{N}-1} <\\varepsilon$  if $n$ is even, and 
    2. \n
  2. $\\left| \\frac{1}{\\sqrt[\\var{m}]{n-1}+1}\\right| \\leq \\frac{1}{\\sqrt[\\var{m}]{N}+1} <\\varepsilon$  if $n$ is odd. 
    3. \n
\n

Then by alternating series test,  $\\displaystyle\\sum_{n=2}^\\infty \\left( \\frac{1}{\\sqrt[\\var{m}]{n}-1} - \\frac{1}{\\sqrt[\\var{m}]{n}+1}\\right)$ converges

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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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