What is lim n → ∞ n ( ∑ k = 1 n 1 n 2 + k − 1 ) ?

Question

What is       lim          n      →      ∞        n  (      ∑          k      =      1              n            1                  n        2            +      k        −  1  ) ?

Ashley Parks · Accepted Answer

We can rewrite the limit as      lim          n      →      ∞            ∑          k      =      1        n        n                  n        2            +      k        −  n  =      lim          n      →      ∞            ∑          k      =      1        n              n      −                        n          2                +        k                            n        2            +      k        =      lim          n      →      ∞            ∑          k      =      1        n              1      −              1        +                  k                      n            2                                      1      +              k                  n          2                    Since the numerator approaches zero we will rationalize and continue to simplify      lim          n      →      ∞            ∑          k      =      1        n              −              k                  n          2                                    1        +                  k                      n            2                              +      1      +              k                  n          2                      =      lim          n      →      ∞            ∑          k      =      1        n              −              k        n                            1        +                  k                      n            2                              +      1      +              k                  n          2                      ⋅      1    n  We can sandwich this complicated expression with two other limits      lim          n      →      ∞        −      1    2        ∑          k      =      1        n        k    n    ⋅      1    n    &amp;lt;      lim          n      →      ∞            ∑          k      =      1        n              −              k        n                            1        +                  k                      n            2                              +      1      +              k                  n          2                      ⋅      1    n    &amp;lt;      lim          n      →      ∞                  −      1                      1        +                  1          n                    +      1      +              1        n                  ∑          k      =      1        n        k    n    ⋅      1    n  Both of which approach the Riemann sum  ⟶  −      1    2        ∫    0    1    x    d  x  =  −      1    4  Thus the limit is                               −                      1            4                               by squeeze theorem.

malalawak44 · Accepted Answer

We have  n      (          ∑              k        =        1            n                      1                              n            2                    +          k                      −    1    )    =  n      ∑          k      =      1        n        (                  1                              n            2                    +          k                      −                  1        n              )    =  n      ∑          k      =      1        n                      n        −                              n            2                    +          k                            n                              n            2                    +          k                        =      ∑          k      =      1        n                      −        k                                          n            2                    +          k                          (          n          +                                    n              2                        +            k                    )                    and, for   1  ≤  k  ≤  n            1              (        n        +        1        )        (        2        n        +        1        )              ≤            1                                    n            2                    +          k                          (          n          +                                    n              2                        +            k                    )                      ≤            1              n        (        2        n        )            while       ∑          k      =      1        n    k  =  n  (  n  +  1  )      /    2. Thus (momentarily dropping the minus sign in the numerator   −  k) we have            n              2        (        n        +        1        )        (        2        n        +        1        )              ≤      ∑          k      =      1        n              k                                    n            2                    +          k                          (          n          +                                    n              2                        +            k                    )                      ≤                    (        n        +        1        )                    4        n            so by the squeeze theorem (and putting the minus sign back in), we have      lim          n      →      ∞        n      (          ∑              k        =        1            n                      1                              n            2                    +          k                      −    1    )    =  −            1      4

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