Using either the Direct or Limit Comparison Tests, determine if ∑ n = 2 ∞ 1 n n 2 − 1 is convergent or divergent.

Question

Using either the Direct or Limit Comparison Tests, determine if       ∑          n      =      2              ∞            1          n                        n          2                −        1             is convergent or divergent.

Milton Gilmore · Accepted Answer

Try comparing your series to      ∑          n      =      2        ∞              1              n        2                  a    n    =            1              n                              n            2                    −          1                      ∼            1              n                              n            2                                =            1              n        ⋅        n              =            1              n        2              =      t    n  The limit comparison text will work very nicely here.                              lim                      n            →            ∞                                                a            n                                t            n                                              =                  lim                      n            →            ∞                                    1                      n                                          n                2                            −              1                                      ×                              n            2                    1                                                  =                  lim                      n            →            ∞                                    n                                    n              2                        −            1                                                            =        1

Jorge Schmitt · Accepted Answer

Let       x    n    =            1              n                              n            2                    −          1                      ≤          1              n                              n            2                    −          2          n          +          1                      =          1              n        (        n        −        1        )              =          1              n        −        1              −          1      n      So we have       y    m    =            ∑              n        =        2                    m                    x      n        ≤          ∑              n        =        1                    m                    1              n        −        1              −          1      n        =    1    −          1      m        &amp;lt;    1  And as             y      m        =          ∑              n        =        2                    m                    x      n       is increasing and bounded so it must be converge.

Using either the Direct or Limit Comparison Tests, determine if sum_(n=2)^infty 1/(n sqrt(n^2-1)) is convergent or divergent.

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