Convergence of ∑ n = 1 ∞ 1 ( n 2 / β ( n 2 β + 1 ) ) 1 / 5

Question

Convergence of       ∑          n      =      1        ∞        1          (              n                  2                      /                    β                    (              n                  2          β                    +      1      )              )                  1                      /                    5

stormsteghj · Accepted Answer

Notice that :      1          (              n                              2            β                              (              n                  2          β                    +      1      )              )                              1            5                                ∼      1          n                        2                      5            β                          +                              2            β                    5                    We deduce that the series :  ∑      1          (              n                              2            β                              (              n                  2          β                    +      1      )              )                              1            5                              converges if, and only if, :      2          5      β        +            2      β        5    &amp;gt;  1if, and only if, :  2      β    2    −  5  β  +  2  &amp;gt;  0if, and only if, :  (  2  β  −  1  )  (  β  −  2  )  &amp;gt;  0if, and only if, :  β  &amp;lt;            1      2         or   β  &amp;gt;  2

Aidyn Cox · Accepted Answer

Clearly, the series diverges for   β  &amp;lt;  0 since as   n  →  ∞, we have      a    n    =      1          [              n                              2            β                              (              n                  2          β                    +      1      )              ]                              1            5                                =            n                        2                      5                          |                        β                          |                                                  (              n                  −          2                      |                    β                      |                              +      1              )                              1            5                                →  +  ∞We also have      a    n    =      1          [              n                              2            β                              (              n                  2          β                    +      1      )              ]                              1            5                                &amp;lt;      b    n    =      1          n                        2                      5            β                          +                              2            β                    5                      ,and       2          5      β        +            2      β        5    &amp;gt;  1 if   p  (  β  )  =  2      β    2    −  5  β  +  2  &amp;gt;  0. The roots of the quadratic polynomial p are       1    2   and 2 and it is easy to check that   p  (  β  )  &amp;gt;  0 when   β  &amp;lt;      1    2  . Hence,   ∑      b    n   is a convergent p-series and the series in question   ∑      a    n   converges by the comparison test for   0  &amp;lt;  β  &amp;lt;      1    2   and   β  &amp;gt;  2For any   β  &amp;gt;  0 and all sufficiently large n, we have       n          2      β        &amp;gt;  1 and      a    n    =      1          [              n                              2            β                              (              n                  2          β                    +      1      )              ]                              1            5                                &amp;gt;      b    n    =      1          [              n                              2            β                              (      2              n                  2          β                    )              ]                              1            5                                =      1                  2                              1            5                                      n                              2                          5              β                                +                                    2              β                        5                              Inspection of   p  (  β  ) from the previous analysis shows that   p  (  β  )  ⩽  0 and       n          2      β        ⩽  1 for       1    2    ⩽  β  ⩽  2. Hence,   ∑      b    n   is a divergent p-series and, by comparison, the series   ∑      a    n   diverges for   β in this range.

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