Prove that this series diverges or converges: ∑ n = 1 ∞ 2 n + n 2 n + n 3

Question

Prove that this series diverges or converges:       ∑          n      =      1              ∞                          2        n            +      n                      2        n            +              n        3

toriannucz · Accepted Answer

Focus on the terms of the series                    2        n            +      n                      2        n            +              n        3              =            2      n                      2        n            +              n        3              +      n                  2        n            +              n        3                  lim          n      →      ∞                  [                  2      n                      2        n            +              n        3                        ]        =  1 and       lim          n      →      ∞                  [            n                  2        n            +              n        3                        ]        =  0so       lim          n      →      ∞                  [                          2        n            +      n                      2        n            +              n        3                        ]        =  1

Willow Pratt · Accepted Answer

a    n    =                    2        n            +      n                      2        n            +              n        3              =            1      +              2                  −          n                    n              1      +              2                  −          n                            n        3              log  ⁡  (      a    n    )  ∼      2          −      n        n  −      2          −      n            n    3    =      2          −      n        n  (  1  −      n    2    )  log  ⁡  (      a          n      +      1        )  −  log  ⁡  (      a    n    )  ∼      2          −      (      n      +      1      )        (  n  −  4  )  n  (  n  +  1  )  &amp;gt;  0    ∀  n  &amp;gt;  4            a              n        +        1                    a      n        =      e          log      ⁡      (              a                  n          +          1                    )      −      log      ⁡      (              a        n            )        &amp;gt;  1If you use   n  =  100, using the exact definition of       a    n  , the exact value is            a              21                    a              20              =      184651888304    184064687179    =  1.003190while the approximation gives            a              21                    a              20              ∼      e          105              /            32768        =  1.003209

Prove that this series diverges or converges: <munderover> ∑ <mrow class

Answered question

Answer & Explanation

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