Finding lim n → ∞ 1 + 2 2 + 3 3 + 4 4 + ⋯ + n n n n Finding lim n → ∞ 1 + 2 2 + 3 3 + 4 4 + ⋯ + n n n n Attempt: lim n → ∞ [ 1 n n + 2 2 n n + 3 3 n n + ⋯ ⋯ + n n n n ] = 1because all terms are approaching to zero except last termsbut answer is not 1 , could some help me to solve it , thanks

Question

Finding       lim          n      →      ∞                  1      +              2        2            +              3        3            +              4        4            +      ⋯      +              n        n                    n      n      Finding      lim          n      →      ∞                  1      +              2        2            +              3        3            +              4        4            +      ⋯      +              n        n                    n      n      Attempt:      lim          n      →      ∞                  [            1          n      n        +            2      2              n      n        +            3      3              n      n        +  ⋯  ⋯  +            n      n              n      n                  ]        =  1because all terms are approaching to zero except last termsbut answer is not   1 , could some help me to solve it , thanks

Giovanna Erickson · Accepted Answer

Bounding by a geometric series,                                                n            n                                n            n                          +                              (            n            −            1                          )                              n                −                1                                                          n            n                          +                              (            n            −            2                          )                              n                −                2                                                          n            n                          +        ⋯        +                              1            1                                n            n                                                            ≤        1        +                  1          n                +                  1                      n            2                          +                  1                      n            3                          +        ⋯                                          =                  n                      n            −            1                              Since the sum is obviously always   ≥  1, and   ≤      n          n      −      1      , the Squeeze Theorem says that the limit is   1

Jamison Rios · Accepted Answer

By Stolz      lim          n      →      ∞                  1      +              2        2            +              3        3            +              4        4            +      ⋯      +              n        n                    n      n        =      lim          n      →      ∞                  n      n                      n        n            −      (      n      −      1              )                  n          −          1                      =  1

Finding lim_(n->oo)(1+2^2+3^3+4^4+...+n^n)/(n^n)

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